linear solutions 2.3
TRANSCRIPT
Friedberg, Insel, and Spence
Linear algebra, 4th ed.
SOLUTIONS REFERENCE
Michael L. Baker<[email protected]>UNIVERSITY OF WATERLOO
January 23, 2011
Preface
The aim of this document is to serve as a reference of problems and solutions from the fourth editionof “Linear Algebra” by Friedberg, Insel and Spence. Originally, I had intended the document tobe used only by a student who was well-acquainted with linear algebra. However, as the documentevolved, I found myself including an increasing number of problems. Therefore, I believe thedocument should be quite comprehensive once it is complete.
I do these problems because I am interested in mathematics and consider this kind of thing to befun. I give no guarantee that any of my solutions are the “best” way to approach the correspondingproblems. If you find any errors (regardless of subtlety) in the document, or you have differentor more elegant ways to approach something, then I urge you to contact me at the e-mail addresssupplied above.
This document was started on July 4, 2010. By the end of August, I expect to have covered up tothe end of Chapter 5, which corresponds to the end of MATH 146, “Linear Algebra 1 (AdvancedLevel)” at the University of Waterloo. This document is currently a work in progress.
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Contents
1 Vector Spaces 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Linear Combinations and Systems of Linear Equations . . . . . . . . . . . . . . . . . 101.5 Linear Dependence and Linear Independence . . . . . . . . . . . . . . . . . . . . . . 161.6 Bases and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.7 Maximal Linearly Independent Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2 Linear Transformations and Matrices 302.1 Linear Transformations, Null spaces, and Ranges . . . . . . . . . . . . . . . . . . . . 302.2 The Matrix Representation of a Linear Transformation . . . . . . . . . . . . . . . . . 342.3 Composition of Linear Transformations and Matrix Multiplication . . . . . . . . . . 36
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1 Vector Spaces
1.1 Introduction
Section 1.1 consists of an introductory, geometrically intuitive treatment of vectors (more specifi-cally, Euclidean vectors). The solutions to the exercises from this section are very basic and as suchhave not been included in this document.
1.2 Vector Spaces
Section 1.2 introduces an algebraic structure known as a vector space over a field, which is then usedto provide a more abstract notion of a vector (namely, as an element of such an algebraic structure).Matrices and n-tuples are also introduced. Some elementary theorems are stated and proved, suchas the Cancellation Law for Vector Addition (Theorem 1.1), in addition to a few uniqueness resultsconcerning additive identities and additive inverses.
8. In any vector space V, show that (a + b)(x + y) = ax + ay + bx + by for any x, y ∈ V and anya, b ∈ F .
Solution. Noting that (a + b) ∈ F , we have
(a + b)(x + y) = (a + b)x + (a + b)y (by VS 7)= ax + bx + ay + by (by VS 8)= ax + ay + bx + by (by VS 1)
as required.
9. Prove Corollaries 1 and 2 [uniqueness of additive identities and additive inverses] of Theorem1.1 and Theorem 1.2(c) [a0 = 0 for each a ∈ F ].
Solution. First, let 0, 0′ ∈ V be additive identities, and let x ∈ V. We have, by VS 3,
x = x + 0 = x + 0′
Whereupon VS 1 yields0 + x = 0′ + x
By Theorem 1.1 [Cancellation Law for Vector Addition], we obtain 0 = 0′, proving that theadditive identity is unique.
Second, let x ∈ V and y, y′ ∈ V be such that x + y = x + y′ = 0. Then we obtain, by VS 1,
y + x = y′ + x
By Theorem 1.1 [Cancellation Law for Vector Addition], we obtain y = y′, proving that eachadditive inverse is unique.
Third, let a ∈ F . We have, by VS 3 and VS 7,
a0 = a(0 + 0) = a0 + a0
By Theorem 1.1 [Cancellation Law for Vector Addition], we obtain a0 = 0, as required.
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10. Let V denote the set of all differentiable real-valued functions defined on the real line. Provethat V is a vector space with the operations of addition and scalar multiplication defined inExample 3.
Solution. Let f, g, h ∈ V and c, d, s ∈ R. Due to rules taught in elementary calculus, wehave f + g ∈ V as well as cf ∈ V for all f, g ∈ V and c ∈ R. Now, note that by thecommutativity of addition in R, we have f + g = f(s) + g(s) = g(s) + f(s) = g + f forall s. Therefore, V satisfies VS 1. Next, by the associativity of addition in R, we have(f + g) + h = (f(s) + g(s)) + h(s) = f(s) + (g(s) + h(s)) = f + (g + h) for all s. Therefore,V satisfies VS 2. Next, let 0V(s) = 0F for all s, where 0F is the additive identity of R.Then f + 0V = f(s) + 0F = f(s) = f for all s. Therefore V satisfies VS 3. Now, for eachs we have f(s) ∈ R and so we are guaranteed an additive inverse, say −f(s), such thatf(s) + (−f(s)) = 0F . Now, 0V(s) = 0F for all s also, so we note f + (−f) = 0V for all f .Therefore V satisfies VS 4. Next, let 1F represent R’s multiplicative identity. Then we have1F ·f = 1F ·f(s) = f(s) = f for all s. Therefore, 1F ·f = f for all f . So VS 5 is satisfied. Notealso that (cd)f = (cd) · f(s) = c · (df(s)) = c(df) for all s. This is because multiplication in Ris associative. So VS 6 is satisfied. We also have c(f +g) = c ·(f(s)+g(s)) = c ·f(s)+c ·g(s) =cf +cg. This is because multiplication in R is distributive (over addition). So VS 7 is satisfied.Finally, note (c + d)f = (c + d) · f(s) = c · f(s) + d · f(s) = cf + df . This is again becausemultiplication in R is distributive (over addition). So VS 8 is satisfied. This completes theproof that V, together with addition and scalar multiplication as defined in Example 3, is avector space.
22. How many matrices are there in the vector space Mm×n(Z2)?
Solution. There are 2mn vectors in this vector space.
1.3 Subspaces
Section 1.3 examines subsets of vector spaces, and defines a special type of subset, known as asubspace, which is a subset of a vector space that can be considered a vector space in its own right(only certain subsets satisfy this property). A certain theorem that is referred to as the subspace testis stated and proven, which provides a fast way of checking whether a given subset is a subspace.More concepts relating to matrices are introduced as well. The section closes with a proof that theintersection of two subspaces is itself a subspace.
3. Prove that (aA + bB)t = aAt + bBt for any A,B ∈ Mm×n(F ) and any a, b ∈ F .
Solution. We have (aA)ij = a ·Aij , (bB)ij = b ·Bij , so (aA + bB)ij = a ·Aij + b ·Bij . So
[(aA + bB)t]ij = a ·Aji + b ·Bji
Now, (At)ij = Aji, (Bt)ij = Bji, (aAt)ij = a ·Aji, (bBt)ij = b ·Bji. So
[aAt + bBt]ij = a ·Aji + b ·Bji
Therefore (aA + bB)t = aAt + bBt as required.
4. Prove that (At)t = A for each A ∈ Mm×n(F ).
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Solution. We have (At)ij = Aji. Thus [(At)t]ij = (At)ji = Aij , so that (At)t = A asrequired.
5. Prove that A + At is symmetric for any square matrix A.
Solution. We have, from exercises 3 and 4:
(A + At) = At + A
= At + (At)t
= (A + At)t
6. Prove that tr(aA + bB) = a tr(A) + b tr(B).
Solution. We have
tr(aA + bB) =n∑
i=1
(aA + bB)ii
=n∑
i=1
(aAii + bBii)
= a
n∑i=1
Aii + b
n∑i=1
Bii
= a · tr(A) + b · tr(B)
19. Let W1 and W2 be subspaces of a vector space V. Prove that W1 ∪W2 is a subspace of V ifand only if W1 ⊆ W2 or W2 ⊆ W1.
Solution. To prove the first direction, assume W1 ∪W2 is a subspace of V. Assume also thatW1 6⊆ W2. Then we are guaranteed some a ∈ W1 such that a /∈ W2. Choose an arbitraryb ∈ W2, and consider the sum a + b. Clearly (a + b) ∈ W1 ∪ W2, and so (a + b) ∈ W1 or(a + b) ∈ W2. Assume that (a + b) ∈ W1. Then, since W1 and W2 are subspaces, we have(a + b) − a = b ∈ W1. The other possibility is that (a + b) ∈ W2, in which case we obtain(a + b) − b = a ∈ W2. However, this is a contradiction, and so we conclude that b ∈ W1,implying W2 ⊆ W1.
To prove the second direction, assume W1 ⊆ W2 or W2 ⊆ W1. Then clearly the union W1∪W2
will be the larger of these two subspaces, and therefore W1 ∪W2 is a subspace.
20. Prove that if W is a subspace of a vector space V and w1, w2, . . . , wn are in W , thena1w1 + a2w2 + . . . + anwn ∈ W for any scalars a1, a2, . . . , an.
Solution. For each i such that 1 ≤ i ≤ n, we have aiwi ∈ W due to W’s closure un-der scalar multiplication. We also know that aiwi + ai+1wi+1 ∈ W for 1 ≤ i ≤ n − 1,since W is closed under addition. An inductive argument can then be used to show thata1w1 + a2w2 + . . . + anwn ∈ W.
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Definition. If S1 and S2 are nonempty subsets of a vector space V, then the sum of S1 and S2,denoted S1 + S2, is the set {x + y : x ∈ S1 and y ∈ S2}.
Definition. A vector space V is called the direct sum of W1 and W2 if W1 and W2 are subspacesof V such that W1 ∩W2 = {0} and W1 + W2 = V . We denote that V is the direct sum of W1 andW2 by writing V = W1 ⊕W2.
23. Let W1 and W2 be subspaces of a vector space V.
(a) Prove that W1 + W2 is a subspace of V that contains both W1 and W2.
Solution. We have that W1 + W2 = {x + y : x ∈ W1 and y ∈ W2}. Since W1,W2 aresubspaces, they both contain the additive identity 0 from V. Therefore, 0 + 0 = 0 ∈(W1 + W2). Next, let a, b ∈ (W1 + W2). Then we have a = xa + ya and b = xb + yb forxa, xb ∈ W1, ya, yb ∈ W2, and so a + b = (xa + ya) + (xb + yb) = (xa + xb) + (ya + yb).Since W1,W2 are subspaces, we have (xa + xb) ∈ W1 and (ya + yb) ∈ W2, and thereforea + b ∈ (W1 + W2). Now let c ∈ F and consider ca = c(xa + ya) = (cxa) + (cya).Since W1,W2 are subspaces, we again have cxa ∈ W1 and cya ∈ W2, and thereforeca ∈ (W1 + W2). We conclude that W1 + W2 is a subspace, by the subspace test. Toshow that W1 ⊆ W1 + W2, simply consider any element u ∈ W1. Since 0 ∈ W2, we canwrite u = u + 0 ∈ W1 + W2 by the definition of W1 + W2, and so W1 ⊆ W1 + W2. Toshow that W2 ⊆ W1 + W2, note 0 ∈ W1 also, and argue similarly.
(b) Prove that any subspace that contains both W1 and W2 must also contain W1 + W2.
Solution. Assume W is some subspace of V, and W1 ⊆ W and W2 ⊆ W. Let a ∈(W1 + W2). Then a = x + y for x ∈ W1 and y ∈ W2. However, because of the above,we obtain x ∈ W and y ∈ W, and since W is a subspace and closed under addition, weobtain a ∈ W. This shows that W1 + W2 ⊆ W.
24. Show that Fn is the direct sum of the subspaces
W1 = {(a1, a2, . . . , an) ∈ Fn : an = 0}
andW2 = {(a1, a2, . . . , an) ∈ Fn : a1 = a2 = . . . = an−1 = 0}.
Solution. First, let’s examine W1 ∩ W2. The definitions of W1 and W2 tell us that if(a1, a2, . . . , an) ∈ W1 ∩W2, then a1 = a2 = . . . = an = 0. This clearly only holds for the n-tuple (0, 0, . . . , 0). Next, let x = (x1, x2, . . . , xn) ∈ Fn. Then we can clearly write x = w1 +w2
for some w1 ∈ W1 and w2 ∈ W2; more specifically, where w1 = (x1, x2, . . . , xn−1, 0) andw2 = (0, 0, . . . , 0, xn). So Fn ⊆ W1 + W2. It is not hard to demonstrate that W1 + W2 ⊆ Fn
to prove that Fn = W1 + W2, proving that Fn is by definition the direct sum of W1 and W2.
31. Let W be a subspace of a vector space V over a field F . For any v ∈ V the set {v} + W ={v + w : w ∈ W} is called the coset of W containing v. It is customary to denote this cosetby v + W rather than {v}+ W.
(a) Prove that v + W is a subspace of V if and only if v ∈ W.
Solution. To prove the first direction, assume v + W is a subspace of V. Clearly,
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v ∈ v + W (since 0 ∈ W). Since v + W is a subspace of V, we also have (1 + 1)v ∈ v + W.Assume, for the sake of contradiction, that v /∈ W. This implies v+v = (1+1)v /∈ v+W.This contradicts the fact that v + W is a subspace of V. Therefore v ∈ W.
To prove the second direction, assume v ∈ W. Since W is a vector space, −v ∈ W also.Therefore v + (−v) = 0 ∈ v + W. Now, let x, y ∈ v + W. Then x = v + wx for somewx ∈ W and y = v + wy for some wy ∈ W. Consider x + y = (v + wx) + (v + wy). Thiscan be expressed as x + y = v + (wx + wy + v). Since W is a vector space, it is closedunder addition, and so (wx + wy + v) ∈ W. The definition of v + W then yields that(x+y) ∈ v+W, thereby proving (v+W)’s closure under addition. Now, let c ∈ F . Thencx = c(v + wx) = cv + cwx. By the closure of W under scalar multiplication, cwx ∈ W.We may write cx = v + (cwx + (c− 1)v). We clearly have cx ∈ v + W, thereby proving(v + W)’s closure under scalar multiplication. This proves v + W is a subspace of V bythe subspace test.
(b) Prove that v1 + W = v2 + W if and only if v1 − v2 ∈ W.
Solution. To prove the first direction, assume v1 + W = v2 + W. Let x ∈ v1 + W. Thenx = v1 + wx for some wx ∈ W. Since v1 + W = v2 + W, there exists some wy ∈ W suchthat x = v1 + wx = v2 + wy. Note that v1 − v2 = wy − wx. W is a subspace and so(wy − wx) ∈ W, thereby proving that (v1 − v2) ∈ W.
To prove the second direction, assume v1 − v2 ∈ W. Let x ∈ v1 + W. Then x = v1 + wx
for some wx ∈ W. Let y = v2 + wy, where wy = (v1 − v2) + wx. Clearly y = v2 + wy =v2 + (v1 − v2) + wx = v1 + wx = x. So we have proved x ∈ v2 + W, thereby proving thatv1 + W ⊆ v2 + W. A similar argument proves that v2 + W ⊆ v1 + W (by using the factthat v2 − v1 ∈ W rather than v1 − v2 ∈ W). Therefore v1 + W = v2 + W.
(c) Addition and scalar multiplication by scalars of F can be defined in the collection S ={v + W : v ∈ V} of all cosets of W as follows:
(v1 + W) + (v2 + W) = (v1 + v2) + W
for all v1, v2 ∈ V anda(v + W) = av + W
for all v ∈ V and a ∈ F . Prove that the preceding operations are well defined; that is,show that if v1 + W = v′1 + W and v2 + W = v′2 + W, then
(v1 + W) + (v2 + W) = (v′1 + W) + (v′2 + W)
anda(v1 + W) = a(v′1 + W)
for all a ∈ F .
Solution. First, we prove the well-definedness of addition. Since v1 + W = v′1 + Wand v2 + W = v′2 + W, we have v1 − v′1 ∈ W, and v2 − v′2 ∈ W by (c). Let A =(v1 + W) + (v2 + W) = (v1 + v2) + W and B = (v′1 + W) + (v′2 + W) = (v′1 + v′2) + W. Leta ∈ A. Then a = (v1+v2)+wa for some wa ∈ W. Let wb = wa+(v1−v′1)+(v2−v′2). Since
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W is a subspace and is closed under addition, we have wb ∈ W. Let b = (v′1 + v′2) + wb.Now we have
b = (v′1 + v′2) + wb
= (v′1 + v′2) + (wa + (v1 − v′1) + (v2 − v′2))= (v1 + v2) + (v′1 − v′1) + (v′2 − v′2) + wa
= (v1 + v2) + wa
= a
Thus, a = b and since b ∈ B, we have that a ∈ B as well, proving A ⊆ B. A similarargument proves that B ⊆ A (by using v′1 − v1 ∈ W and v′2 − v2 ∈ W rather thanv1 − v′1 ∈ W and v2 − v′2 ∈ W). Therefore A = B, and so addition is well-defined.
Second, we prove the well-definedness of scalar multiplication. Let c ∈ F . Since v1 +W = v′1 + W, we have v1 − v′1 ∈ W by (c). Let A = c(v1 + W) = cv1 + W andB = c(v′1 + W) = cv′1 + W. Let a ∈ A. Then a = cv1 + wa for some wa ∈ W. Sincev1 − v′1 ∈ W, we have that c(v1 − v′1) = cv1 − cv′1 ∈ W. Let wb = wa + (cv1 − cv′1). SinceW is a subspace and is closed under addition, we have wb ∈ W. Let b = cv′1 + wb ∈ B.Now we have
b = cv′1 + wb
= cv′1 + (wa + (cv1 − cv′1))= cv1 + (cv′1 − cv′1) + wa
= cv1 + wa
= a
Thus, a = b and since b ∈ B, we have that a ∈ B as well, proving A ⊆ B. A similarargument proves that B ⊆ A (by using cv′1 − cv1 ∈ W rather than cv1 − cv′1 ∈ W).Therefore A = B, and so scalar multiplication is well-defined.
(d) Prove that S is a vector space with the operations defined in (c). This vector space iscalled the quotient space of V modulo W and is denoted by V/W.
Solution. To verify VS 1, let x, y ∈ S. Then x = vx+W, y = vy+W for some vx, vy ∈ V.We have x+y = (vx +W)+(vy +W) = (vx +vy)+W, and y+x = (vy +W)+(vx +W) =(vy + vx) + W. Since addition in V is commutative, however, we have vx + vy = vy + vx,thereby demonstrating that x + y = y + x.
To verify VS 2, let x, y, z ∈ S. Then x = vx + W, y = vy + W, z = vz + W for somevx, vy, vz ∈ V. We have
(x + y) + z = ((vx + W) + (vy + W)) + (vz + W)= ((vx + vy) + W) + (vz + W)= ((vx + vy) + vz) + W
= (vx + (vy + vz)) + W
= (vx + W) + ((vy + W) + (vz + W))= x + (y + z)
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This is due to the fact that addition in V is associative.
To verify VS 3, let 0′ = (0 + W). Clearly 0′ ∈ S, since 0 ∈ V (this must hold, since V isa vector space). Now, let s ∈ S. Then s = vs + W for some vs ∈ V. We have
0′ + s = (0 + W) + (vs + W) = (0 + vs) + W = vs + W = s
as needed.
To verify VS 4, let x ∈ S. Then x = vx +W for some vx ∈ V. Let y = −vx +W. Clearly,−vx ∈ V since V is a vector space. Then x+y = (vx +W)+(−vx +W) = (vx−vx)+W =0 + W, which is the zero vector of S described in VS 3.
To verify VS 5, let s ∈ S. Then s = vs + W for some vs ∈ V. We have
1s = 1(vs + W)= (1vs) + W
= vs + W
= s
This is due to the fact that since V is a vector space, 1v = v for all v ∈ V.
To verify VS 6, let a, b ∈ F and s ∈ S. Then s = vs + W for some vs ∈ V. We have
(ab)s = (ab)(vs + W)= (ab)vs + W
= a(bvs) + W
= a(bvs + W)= a(bs)
This is due to the fact that as a vector space, V satisfies VS 6.
To verify VS 7, let a ∈ F and x, y ∈ S. Then x = vx + W and y = vy + W for somevx, vy ∈ V. We have
a(x + y) = a[(vx + W) + (vy + W)]= a[(vx + vy) + W]= [a(vx + vy)] + W
= [avx + avy] + W
= (avx + W) + (avy + W)= a(vx + W) + a(vy + W)= ax + ay
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This is due to the fact that as a vector space, V satisfies VS 7.
To verify VS 8, let a, b ∈ F and s ∈ S. Then s = vs + W for some vs ∈ V. We have
(a + b)s = (a + b)[vs + W]= [(a + b)vs] + W
= [avs + bvs] + W
= (avs + W) + (bvs + W)= a(vs + W) + b(vs + W)= as + bs
This is due to the fact that as a vector space, V satisfies VS 8.
Thus, S = V/W is a vector space, since we have verified all the vector space axioms.
1.4 Linear Combinations and Systems of Linear Equations
Section 1.4 discusses the notion of linear combination and also provides a method for solving systemsof linear equations which will later serve as the foundation for the row reduction of matrices. Thespan of a subset is introduced, in addition to the notion of a generating subset. Some geometricinterpretations of the concepts are given.
2. Solve the following systems of linear equations by the method introduced in this section.
(a)
2x1 − 2x2 − 3x3 = −2 (1.4.1)3x1 − 3x2 − 2x3 + 5x4 = 7 (1.4.2)x1 − x2 − 2x3 − x4 = −3 (1.4.3)
Solution. Begin by replacing equation (1.4.2) with (1.4.2)− 3 · (1.4.3) to obtain 4x3 +8x4 = 16. Divide equation (1.4.2) by 4 to obtain x3 + 2x4 = 4. Now replace equation(1.4.1) with (1.4.1)− 2 · (1.4.3) to obtain x3 +2x4 = 4 also. Switch equations (1.4.3) and(1.4.1). We obtain
x1 − x2 − 2x3 − x4 = −3x3 + 2x4 = 4
Which gives us
x3 = 4− 2x4
x1 = x2 − 3x4 + 5
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Therefore, the solution set is the set
S = {(x2 − 3x4 + 5, x2,−2x4 + 4, x4) : x2, x4 ∈ R}
Which can also be expressed, by letting x2 = s and x4 = t, as
S = {s(1, 1, 0, 0) + t(−3, 0,−2, 1) + (5, 0, 4, 0) : s, t ∈ R}
simply by examining the coefficients of the different xi, in addition to the constants, inthe first representation of the set. This solution set represents a plane in R3 becausethere are two parameters (s and t).
(b)
3x1 − 7x2 + 4x3 = 10 (1.4.4)x1 − 2x2 + x3 = 3 (1.4.5)
2x1 − x2 − 2x3 = 6 (1.4.6)
Solution. Begin by replacing equation (1.4.4) with (1.4.4)− 3 · (1.4.5) to obtain −x2 +x3 = 1. Next, replace equation (1.4.6) with (1.4.6)− 2 · (1.4.5) to obtain 3x2 − 4x3 = 0.Now replace equation (1.4.6) with (1.4.6)+3·(1.4.4) to obtain −x3 = 3. Replace equation(1.4.6) with −(1.4.6) to obtain x3 = −3. Replace equation (1.4.4) with (1.4.4)− (1.4.6)to obtain −x2 = 4. Replace equation (1.4.4) with −(1.4.4) to obtain x2 = −4. Replaceequation (1.4.5) with (1.4.5)+2 ·(1.4.4) to obtain x1+x3 = −5. Replace equation (1.4.5)with (1.4.5)− (1.4.6) to obtain x1 = −2. Therefore, the solution set is the set
S = {(−2,−4,−3)}
which is a single point in R3.
(c)
x1 + 2x2 − x3 + x4 = 5 (1.4.7)x1 + 4x2 − 3x3 − 3x4 = 6 (1.4.8)
2x1 + 3x2 − x3 + 4x4 = 8 (1.4.9)
Solution. Begin by replacing equation (1.4.8) with (1.4.8)−(1.4.7) to obtain 2x2−2x3−4x4 = 1. Next, replace equation (1.4.9) with (1.4.9)− 2 · (1.4.7) obtain −x2 +x3 +2x4 =−2. Now, replace equation (1.4.9) with (1.4.9) + 1
2 · (1.4.8), to obtain 0 = − 52 . This
shows that the system is inconsistent and therefore our solution set is the empty set,
S = ∅
3. For each of the following lists of vectors in R3, determine whether the first vector can beexpressed as a linear combination of the other two.
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(a) (−2, 0, 3), (1, 3, 0), (2, 4,−1)
Solution. We need to verify that there exist s, t ∈ R such that
(−2, 0, 3) = s(1, 3, 0) + t(2, 4,−1) = (s + 2t, 3s + 4t,−t)
This yields the following system of equations:
s + 2t = −2 (1.4.10)3s + 4t = 0 (1.4.11)
− t = 3 (1.4.12)
Applying the same procedure as previously, we obtain s = 4 and t = −3. The sys-tem therefore has solutions, and so we conclude (−2, 0, 3) can be expressed as a linearcombination of (1, 3, 0) and (2, 4,−1).
4. For each list of polynomials in P3(R), determine whether the first polynomial can be expressedas a linear combination of the other two.
(b) 4x3 + 2x2 − 6, x3 − 2x2 + 4x + 1, 3x3 − 6x2 + x + 4
Solution. We need to verify that there exist s, t ∈ R such that
4x3 + 2x2 − 6 = s(x3 − 2x2 + 4x + 1) + t(3x3 − 6x2 + x + 4)
= (s + 3t)x3 + (−2s− 6t)x2 + (4s + t)x + (s + 4t)
This yields the following system of equations:
s + 3t = 4 (1.4.13)−2s− 6t = 2 (1.4.14)
4s + t = 0 (1.4.15)s + 4t = −6 (1.4.16)
Applying the same procedure as previously, we obtain −2s − 6t = −8, which is incon-sistent with (1.4.14). The system therefore has no solutions, and so we conclude that4x3 + 2x2 − 6 cannot be expressed as a linear combination of x3 − 2x2 + 4x + 1 and3x3 − 6x2 + x + 4.
10. Show that if
M1 =(
1 00 0
), M2 =
(0 00 1
), M3 =
(0 11 0
),
then the span of {M1,M2,M3} is the set of all symmetric 2× 2 matrices.
Solution. Let A denote the set of all symmetric 2 × 2 matrices. First, let a ∈ A. Then wehave, by the definition of a symmetric matrix,
a =(
x11 nn x22
)
12
We see easily that
x11 ·(
1 00 0
)+ n ·
(0 11 0
)+ x22 ·
(0 00 1
)=(
x11 nn x22
)= a
and hence a ∈ span(S) where we define S = {M1,M2,M3}. This shows that A ⊆ span(S). Asimilar argument demonstrates that span(S) ⊆ A, and this completes the proof.
11. Prove that span({x}) = {ax : a ∈ F} for any vector x in a vector space. Interpret this resultgeometrically in R3.
Solution. Since {x} contains only one vector, then for all a ∈ span({x}) we can write, forsome c ∈ F ,
a = cx
which proves span({x}) ⊆ {ax : a ∈ F}. Clearly, if b ∈ {ax : a ∈ F} then b is a linearcombination of the vectors in {x}, namely, a scalar multiple of x itself, so that {ax : a ∈ F} ⊆span({x}). This proves span({x}) = {ax : a ∈ F}. In R3 this result tells us that the span ofany single vector will be a line passing through the origin.
12. Show that a subset W of a vector space V is a subspace of V if and only if span(W) = W.
Solution. Assume W ⊆ V such that span(W) = W. 0 is in the span of any subset of V dueto the trivial representation. Now let a, b ∈ W. Since span(W) = W, we have a, b ∈ span(W),so we can write
a + b =
(n∑
i=1
aivi
)+
(n∑
i=1
bivi
)=
n∑i=1
(ai + bi)vi
for some a1, . . . , an, b1, . . . , bn ∈ F and v1, . . . , vn ∈ W, whereby we clearly see a+b ∈ span(W),and hence a + b ∈ W. This proves that W is closed under addition. Now consider, for somec ∈ F ,
ca = c
(n∑
i=1
aivi
)=
n∑i=1
(cai)vi
which demonstrates that ca ∈ span(W) and hence ca ∈ W. This proves that W is closed underscalar multiplication. By the subspace test, we have that W is a subspace.
Assume next that W ⊆ V with W a subspace of V. Since W is a subspace of V, clearly0 ∈ W, and W is closed under addition and scalar multiplication (in other words, linearcombination). Any vector a ∈ W can be trivially represented as a linear combination ofvectors in W (viz., itself multiplied by the scalar 1 from F ) and so we have a ∈ span(W),proving that W ⊆ span(W). Now, let a ∈ span(W). Then, by the definition of span(W), we seethat a is a linear combination of vectors in W. By noting first that W is closed under scalarmultiplication and then noting its closure under addition, we clearly have a ∈ W, provingspan(W) ⊆ W, and hence span(W) = W.
13. Show that if S1 and S2 are subsets of a vector space V such that S1 ⊆ S2, then span(S1) ⊆span(S2). In particular, if S1 ⊆ S2 and span(S1) = V, deduce that span(S2) = V.
13
Solution. Assume a ∈ span(S1). Then for a1, a2, . . . , an ∈ F and v1, v2, . . . , vn ∈ S1 we canwrite
a =n∑
i=1
aivi
However, we have S1 ⊆ S2, which implies that for each i, we have vi ∈ S2 also. By thedefinition of the linear combination, then, we obtain that a ∈ span(S2), demonstrating thatspan(S1) ⊆ span(S2). Now assume S1 ⊆ S2 and span(S1) = V. We obtain (from theproposition we just proved) that span(S1) = V ⊆ span(S2). However, each element in S2
is also in V. Since V is closed under vector addition and scalar multiplication, surely everylinear combination of vectors in S2 must be in V. Therefore, span(S2) ⊆ V, which, sinceV ⊆ span(S2) also, shows that span(S2) = V in this case.
14. Show that if S1 and S2 are arbitrary subsets of a vector space V, then span(S1 ∪ S2) =span(S1) + span(S2). (The sum of two subsets is defined in the exercises of Section 1.3.)
Solution. We will define
A = span(S1) + span(S2) = {(a + b) : a ∈ span(S1), b ∈ span(S2)}B = span(S1 ∪ S2)
Assume a ∈ A. Then we have a = u + v for some u ∈ span(S1) and v ∈ span(S2). There-fore, we can write, for some a1, a2, . . . , am, b1, b2, . . . , bn ∈ F and x1, x2, . . . , xm ∈ S1 andy1, y2, . . . , yn ∈ S2:
a =
(m∑
i=1
aixi
)+
(n∑
i=1
biyi
)Since, for each i, we have xi ∈ S1 and hence xi ∈ (S1∪S2), and yi ∈ S2 and hence yi ∈ (S1∪S2),we clearly see that u, v ∈ span(S1 ∪ S2). Since by Theorem 1.5, the span of any subset of avector space V is a subspace of V, it is closed under vector addition, implying (u+v) = a ∈ B,we have obtained A ⊆ B.
Now, assume b ∈ B. Then we can write, for some a1, a2, . . . , an ∈ F and z1, z2, . . . , zn ∈(S1 ∪ S2), the following:
b =n∑
i=1
aizi
Therefore, for each i, we see that zi satisfies either zi ∈ S1 or zi ∈ S2 (or both), for otherwisewe have zi /∈ (S1 ∪ S2). Now, we partition the set of summands as follows:
P = {aizi : zi ∈ S1, 0 ≤ i ≤ n}Q = {aizi : zi ∈ S2, zi /∈ S1, 0 ≤ i ≤ n}
It can easily be proved that P ∪Q yields all the summands in the expression for b and thatP ∩Q = ∅. Then we can write the decomposition
b =n∑
i=1
aizi =∑x∈P
x +∑x∈Q
x
14
We clearly have that (∑
x∈P x) ∈ span(S1) and (∑
x∈Q x) ∈ span(S2) which proves thatb ∈ A, whereupon we obtain B ⊆ A. We have therefore proved that A = B, as required.
15. Let S1 and S2 be subsets of a vector space V. Prove that span(S1∩S2) ⊆ span(S1)∩span(S2).Give an example in which span(S1 ∩ S2) and span(S1)∩ span(S2) are equal and one in whichthey are unequal.
Solution. We will define
A = span(S1 ∩ S2)B = span(S1) ∩ span(S2)
Assume a ∈ A. Then we can write, for some a1, a2, . . . , an ∈ F and v1, v2, . . . , vn,
a =n∑
i=1
aivi
such that for each i, vi satisfies both vi ∈ S1 and vi ∈ S2. Because of this, we see immediatelythat a ∈ span(S1) and a ∈ span(S2), proving that a ∈ (span(S1) ∩ span(S2)). Therefore,A ⊆ B, as required.
Now, let V = R2 and S1 = {(−1, 1)}, S2 = {(1, 1)}. Then S1 ∩ S2 = ∅, so that A =span(S1 ∩ S2) = {0} and B = span(S1) ∩ span(S2) = {0}. This is a case in which A = B.
Next, let V = R2 and S1 = {(2, 0)} and S2 = {(1, 0)}. Then S1 ∩ S2 = ∅, yielding A =span(S1 ∩ S2) = span ∅ = {0}. Now,
span(S1) = {k(2, 0) : k ∈ R}span(S2) = {k(1, 0) : k ∈ R}
It is not hard to show that span(S1) = span(S2), so we conclude B = span(S1) ∩ span(S2) =span(S1). Clearly, span(S1) 6= ∅, so this is a case in which A ( B.
16. Let V be a vector space and S a subset of V with the property that whenever v1, v2, . . . , vn ∈ Sand a1v1 + a2v2 + · · · + anvn = 0, then a1 = a2 = · · · = an = 0. Prove that every vector inthe span of S can be uniquely written as a linear combination of vectors of S.
Solution. Assume v ∈ span(S), and assume also that we can represent v in two ways, likeso, for some a1, a2, . . . , an, b1, b2, . . . , bn ∈ F and v1, v2, . . . , vn ∈ S:
v =n∑
i=1
aivi =n∑
i=1
bivi
Then we clearly obtain
0 = v − v =
(n∑
i=1
aivi
)−
(n∑
i=1
bivi
)=
n∑i=1
(ai − bi)vi
At this point, we use the given property of S, and this gives us that for each i, we have
ai − bi = 0
15
which implies ai = bi. This clearly proves that our two representations of v cannot be distinct,as required.
17. Let W be a subspace of a vector space V. Under what conditions are there only a finitenumber of distinct subsets S of W such that S generates W?
Solution. I believe the following conditions would lead to there only being a finite numberof distinct generating sets for W:
• The case where W = {0}, in which case there would only be the set {0} and ∅ thatgenerate W.
• The case where F is a Galois field such as Z2.
1.5 Linear Dependence and Linear Independence
Section 1.5 discusses the notion of linear dependence, first through the question “Can one of thevectors in this set be represented as a linear combination of the others?” and then reformulatesthis notion in terms of being able to represent the zero vector as a nontrivial linear combination ofvectors in the set. Afterwards, some proofs concerning linear dependence and linear independenceare given.
2. Determine whether the following sets are linearly dependent or linearly independent.
(a){(
1 −3−2 4
),
(−2 6
4 −8
)}in M2×2(R)
Solution. We consider the equation
a1
(1 −3
−2 4
)+ a2
(−2 6
4 −8
)=(
a1 − 2a2 −3a1 + 6a2
−2a1 + 4a2 4a1 − 8a2
)=(
0 00 0
)This yields the following system of equations:
a1 − 2a2 = 0 (1.5.1)−3a1 + 6a2 = 0 (1.5.2)−2a1 + 4a2 = 0 (1.5.3)
4a1 − 8a2 = 0 (1.5.4)
Solving the system with the usual method, we obtain a1 = 2a2. Therefore there areinfinitely many nontrivial solutions and so the set is linearly dependent in M2×2(R).
(b){(
1 −2−1 4
),
(−1 1
2 −4
)}in M2×2(R)
Solution. We consider the equation
a1
(1 −2
−1 4
)+ a2
(−1 1
2 −4
)=(
a1 − a2 −2a1 + a2
−a1 + 2a2 4a1 − 4a2
)=(
0 00 0
)
16
This yields the following system of equations:
a1 − a2 = 0 (1.5.5)−2a1 + a2 = 0 (1.5.6)− a1 + 2a2 = 0 (1.5.7)
4a1 − 4a2 = 0 (1.5.8)
Solving the system with the usual method, we obtain a1 = 0 and a2 = 0 (which provesthat only trivial solutions exist) and so the set is linearly independent in M2×2(R).
4. In Fn, let ej denote the vector whose jth coordinate is 1 and whose other coordinates are 0.Prove that {e1, e2, . . . , en} is linearly independent.
Solution. Upon writing out the system of equations, we see that the system immediatelyyields ai = 0 for all i satisfying 1 ≤ i ≤ n. Therefore, {e1, e2, . . . , en} is linearly independent.
7. Recall from Example 3 in Section 1.3 that the set of diagonal matrices in M2×2(F ) is asubspace. Find a linearly independent subset that generates this subspace.
Solution. Since all diagonal matrices are in the form(a 00 b
)= a
(1 00 0
)+ b
(0 00 1
)We can simply use the set:
S ={(
1 00 0
),
(0 00 1
)}8. Let S = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} be a subset of the vector space F3.
(a) Prove that if F = R, then S is linearly independent.
Solution. Assume F = R. Then we must prove there exist no nontrivial solutionsa1, a2, a3 ∈ R to
a1(1, 1, 0) + a2(1, 0, 1) + a3(0, 1, 1) = (0, 0, 0)
The equation above yields the following system:
a1 + a2 = 0 (1.5.9)a1 + a3 = 0 (1.5.10)a2 + a3 = 0 (1.5.11)
Solving the system with the usual method, we obtain a1 = a2 = a3 = 0.
9. Let u and v be distinct vectors in a vector space V. Show that {u, v} is linearly dependent ifand only if u or v is a multiple of the other.
Solution. First, assume without loss of generality that v = cu for some c ∈ F . We assumec 6= 0, for otherwise v = 0u = 0 which implies that {u, v} is linearly dependent. Also assumeu 6= 0 for the same reason. Then we obtain a1v + a2u = 0, which becomes a1cu + a2u =
17
(a1c + a2)u = 0 by VS 8. Then the equation will hold for any choice of a1, a2 ∈ F such thata1c + a2 = 0. This equation can clearly be satisfied by some choice of a1, a2 ∈ F , not bothzero, because if we let a1 = 1 and a2 = −c (these elements are guaranteed to be in F dueto the field axioms), we obtain a1c + a2 = 1 · c + (−c) = c + (−c) = 0 by the field axioms.Therefore, {u, v} is linearly dependent.
Now, assume {u, v} is linearly dependent. Then there exist a1, a2 ∈ F , not all zero, such thata1v + a2u = 0. Since F is a field, division is possible, allowing us to write
v = −a2
a1u
which signifies v is a multiple of u, as required. Note that we have assumed above a1 6= 0,for otherwise we have a2 6= 0, which from the equation implies u = 0, and we know that 0 is(trivially) a multiple of any vector. This completes the proof.
11. Let S = {u1, u2, . . . , un} be a linearly independent subset of a vector space V over the fieldZ2. How many vectors are there in span(S)? Justify your answer.
Solution. The span of S will consist of every linear combination of vectors in S, or in somesubset of S. The field Z2 only admits two possible scalars, 0 and 1, therefore every subset ofS can be regarded as a linear combination and vice versa. The power set of S (that is, theset of all subsets of S) will contain 2n elements (by a theorem in set theory), since S containsn elements. There is therefore a bijection between the set of all subsets of S and the set ofall linear combinations of vectors in S. For this reason, the cardinality of span(S) must alsobe 2n.
12. Prove Theorem 1.6 [supersets of a linearly dependent set are themselves linearly dependent]and its corollary [subsets of a linearly independent set are themselves linearly independent].
Solution. Let V be a vector space and let S1 ⊆ S2 ⊆ V. Assume S1 is linearly dependent.We wish to show that S2 is linearly dependent. Since S1 is linearly dependent, there exista1, a2, . . . , an ∈ F , not all zero, and v1, v2, . . . , vn ∈ S1 such that a1v1 +a2v2 + . . .+anvn = 0.However, each vi (1 ≤ i ≤ n) is also in S2 since S1 ⊆ S2. For this reason, S2 is also linearlydependent.
To prove the corollary, let V be a vector space and let S1 ⊆ S2 ⊆ V. Assume S2 is linearlyindependent. We wish to show that S1 is linearly independent. To prove this, assume forthe sake of contradiction that S1 is linearly dependent. Then by the argument above, S2 islinearly dependent. Since linear independence and linear dependence are mutually exclusiveproperties, this is a contradiction, and the corollary follows.
13. Let V be a vector space over a field of characteristic not equal to two.
(a) Let u and v be distinct vectors in V. Prove that {u, v} is linearly independent if andonly if {u + v, u− v} is linearly independent.
Solution. First, assume {u, v} is linearly independent. Then the equation a1u+a2v = 0,with a1, a2 ∈ F , holds only if a1 = a2 = 0. Assume, for the sake of contradiction,that {u + v, u − v} is linearly dependent. Then there exist b1, b2 ∈ F , not all zero,such that b1(u + v) + b2(u − v) = 0. Note that this equation can be rewritten as
18
(b1 + b2)u+(b1− b2)v = 0. For the first case, assume b1 + b2 = 0. Then b2 = −b1 and sob1− b2 = b1− (−b1) = b1 + b1 6= 0, since the field’s characteristic is not 2. For the secondcase, assume b1 − b2 = 0. Then b1 = b2 and so b1 + b2 = b1 + b1 6= 0, since the field’scharacteristic is not 2. In either case we obtain two scalars x, y ∈ F , at least one of whichis nonzero, that satisfy the equation xu + yv = 0. This implies that {u, v} is linearlydependent, which is a contradiction. Therefore, {u + v, u− v} is linearly independent.
Now, assume {u + v, u − v} is linearly independent. Then the equation a1(u + v) +a2(u − v) = 0 is satisfied only when a1 = a2 = 0. This equation can be rewritten as(a1 +a2)u+(a1−a2)v = 0. Assume, for the sake of contradiction, that {u, v} is linearlydependent. Then there exist b1, b2 ∈ F , not all zero, such that b1u + b2v = 0. For thefirst case, assume b1 6= 0. Then any scalars a1, a2 ∈ F such that a1 + a2 = b1 will satisfythe equation a1(u + v) + a2(u − v) = 0. Clearly these scalars a1, a2 are not both zero,since otherwise the equation a1 + a2 = b1 could not hold. For the second case, assumeb2 6= 0. Then any scalars a1, a2 ∈ F such that a1 − a2 = b2 will satisfy the equationa1(u + v) + a2(u− v) = 0. Again, we see that at least one of these scalars is nonzero. Ineither case we obtain that {u + v, u− v} is linearly dependent, which is a contradiction.Therefore, {u, v} is linearly independent.
14. Prove that a set S is linearly dependent if and only if S = {0} or if there exist distinct vectorsv, u1, u2, . . . , un in S such that v is a linear combination of u1, u2, . . . , un.
Solution. Assume that S is a linearly dependent subset of a vector space V over a fieldF . Then S is nonempty (for otherwise it cannot be linearly dependent) and contains vectorsv1, . . . , vn. By definition of linear dependence, the equation a1v1 + . . . + anvn = 0 is satisfiedby some a1, . . . , an ∈ F , not all zero. If n = 1, then the equation a1v1 = 0 for some nonzeroa1 ∈ F implies that v1 = 0 and so S = {0}, so assume n ≥ 2. Then there exists k satisfying1 ≤ k ≤ n such that ak 6= 0. Then the equation a1v1 + . . . + akvk + . . . + anvn = 0 canbe rearranged to yield akvk = −a1v1 − . . .− anvn, whereupon we divide both sides by ak toobtain
vk = −(
a1
ak
)v1 − . . .−
(an
ak
)vn
which proves that vk is a linear combination of some finite subset of vectors from S, asrequired.
Next, assume that either S = {0} or there exist v, u1, u2, . . . , un ∈ S such that v is a linearcombination of the vectors u1, . . . , un. If S = {0}, then S is clearly linearly dependent, sincek0 = 0 for any k ∈ F by Theorem 1.2(c) which was proved as exercise 9 of section 1.2. Ifthere exist v, u1, . . . , un ∈ S such that v is a linear combination of u1, . . . , un, then we obtainthe equation
v = a1u1 + . . . + anun
and observing that the coefficient of v is 1, we rearrange the equation to obtain v − a1u1 −. . . − anun = 0 which, being a nontrivial representation of 0 as a linear combination of thevectors in S, implies that S is linearly dependent as required.
15. Let S = {u1, u2, . . . , un} be a finite set of vectors. Prove that S is linearly dependent if andonly if u1 = 0 or uk+1 ∈ span({u1, u2, . . . , uk}) for some k (1 ≤ k ≤ n).
19
Solution. Assume S is linearly dependent. Then by exercise 14, either S = {0} or thereare v, u1, . . . , un ∈ S such that v is a linear combination of S. If S = {0} then u1 = 0 asrequired. Otherwise, we can write uk = v so that uk = a1u1 + a2u2 + . . . + anun for somea1, . . . , an ∈ F . Either k > n and the proof is complete, since the fact that we can write uk =a1u1+a2u2+. . .+anun+0un+1+. . .+0uk−1 clearly implies uk ∈ span({u1, . . . , un, . . . , uk−1})as required. Otherwise we can simply isolate un in the equation and use the same reasoning.
Now, assume either S = {0} or there exist uk+1 ∈ S such that uk+1 ∈ span({u1, u2, . . . , uk})for some k (1 ≤ k ≤ n). If S = {0}, then S is linearly dependent by exercise 14. Otherwise,the fact that uk+1 ∈ span({u1, u2, . . . , uk}) for some k satisfying 1 ≤ k ≤ n implies that thereexists some vector v ∈ S (namely, uk+1) that is a linear combination of some other vectors inS (namely the set {u1, u2, . . . , uk}). Exercise 14 then implies that S is linearly dependent.
16. Prove that a set S of vectors is linearly independent if and only if each finite subset of S islinearly independent.
Solution. Assume S is linearly independent. For each finite subset R ⊆ S, R must be linearlyindependent by exercise 12, as required.
Next, assume that each finite subset R ⊆ S is linearly independent. Assume, for the sakeof contradiction, that S is linearly dependent. Then there exist vectors v1, . . . , vn ∈ S, andscalars a1, . . . , an ∈ F , not all zero, such that a1v1 + . . . + anvn = 0. This implies that somefinite subset R ⊆ S is linearly dependent, which is contradictory to what we assumed. Thiscompletes the proof.
17. Let M be a square upper triangular matrix (as defined in Exercise 12 of Section 1.3) withnonzero diagonal entries. Prove that the columns of M are linearly independent.
Solution. Clearly, Mi,j = 0 whenever i > j and Mi,j 6= 0 whenever i = j. We wish to provethat the equation
a1
M1,1
· · ·Mn,1
+ a2
M1,2
· · ·Mn,2
+ . . . + an
M1,n
· · ·Mn,n
=
0· · ·0
only holds when a1 = . . . = an = 0. This, of course, yields the n equations:
a1M1,1 + a2M1,2 + . . . + anM1,n = 0a1M2,1 + a2M2,2 + . . . + anM2,n = 0
...a1Mn,1 + a2Mn,2 + . . . + anMn,n = 0
We proceed by induction. Let P (k) be the statement that an−k+1 = 0 (1 ≤ k ≤ n). ConsiderP (1). We wish to show that an−1+1 = 0, or in other words an = 0. Looking at the equationn− 1 + 1, or in other words equation n:
a1Mn,1 + a2Mn,2 + . . . + anMn,n = 0
we note that each Mn,i = 0 unless i = n, since this is the only case when n > i does not hold(the matrix is upper triangular). This allows us to remove the first n− 1 summands from the
20
equation, leaving us withanMn,n = 0
which immediately yields an = 0, since Mn,n 6= 0 (the entry lies on the diagonal, so it isnonzero). Therefore P (1) holds. Now, assume P (i) holds for all 1 ≤ i ≤ k. We wish to showthat P (k + 1) holds. In order to do this, we examine equation n − (k + 1) + 1, or in otherwords n− k:
a1Mn−k,1 + a2Mn−k,2 + . . . + anMn−k,n = 0
Since the matrix is upper triangular, Mn−k,i = 0 for all n−k > i. Then the equation becomes
an−kMn−k,n−k + an−k+1Mn−k,n−k+1 + an−k+2Mn−k,n−k+2 + . . . + anMn−k,n = 0
We now use the induction hypothesis, which tells us that an−k+1 = an−k+2 = . . . = an = 0.This causes the equation to reduce again, to
an−kMn−k,n−k = 0
whereupon we note that Mn−k,n−k 6= 0 since it lies on the diagonal. This immediatelyyields that an−k = 0, proving that P (k + 1) is true. By strong induction, this proves thata1 = a2 = . . . = an = 0 as required.
19. Prove that if {A1, A2, . . . , Ak} is a linearly independent subset of Mn×n(F ), then {At1, A
t2, . . . , A
tk}
is also linearly independent.
Solution. Since {A1, A2, . . . , Ak} is linearly independent, the equation
a1A1 + a2A2 + . . . + akAk = 0
holds only for a1 = a2 = . . . = ak = 0. Taking the transpose of both sides and applying anearlier result we obtain
(a1A1 + a2A2 + . . . + akAk)t = 0t
a1At1 + a2A
t2 + . . . + akAt
k = 0
The linear independence of {At1, A
t2, . . . , A
tk} follows.
1.6 Bases and Dimension
Section 1.6 presents, and gives examples of, the concept of a basis, that is, a linearly independentsubset of a vector space that generates the vector space. Theorem 1.8 says that vectors in thevector space can be represented as unique linear combinations of vectors in a subset if and onlyif the subset is a basis. Theorem 1.9 says that every finite generating set for a vector space canbe reduced to a basis for that vector space. Theorem 1.10 (the replacement theorem) gives a veryimportant result: that a linearly independent subset in a vector space has no more elements in itthan a set that generates that vector space, and that a subset of this generator can be unionedwith the independent subset to form a generating set. Corollaries to this theorem are the fact thatevery basis in a finite-dimensional vector space contains the same number of elements, that anyfinite generating set for a vector space V contain at least dim(V) vectors, and that a generating setcontaining exactly dim(V) vectors is a basis. Additionally, any linearly independent subset of Vwith exactly dim(V) vectors is a basis, and every linearly independent subset of V can be extendedto a basis for V.
21
2. Determine which of the following sets are bases for R3.
(a) {(1, 0,−1), (2, 5, 1), (0,−4, 3)}
Solution. By a corollary to the replacement theorem, any linearly independent subsetof 3 vectors generates R3, and so is a basis for R3. Therefore this problem is reducedsimply to determining whether the set is linearly independent, which of course involvessolving the usual homogeneous system of equations (that is, a system of equations whereall the right-hand sides are equal to 0). This has been demonstrated previously in thedocument, and so will not be shown here.
4. Do the polynomials x3−2x2 +1, 4x2−x+3, and 3x−2 generate P3(R)? Justify your answer.
Solution. No. This is a set of 3 polynomials, and the dimension of P3(R) is 4. By acorollary to the replacement theorem, any set which generates a 4-dimensional vector spacemust contain at least 4 vectors.
5. Is {(1, 4,−6), (1, 5, 8), (2, 1, 1), (0, 1, 0)} a linearly independent subset of R3? Justify youranswer.
Solution. No. This set contains 4 vectors, and R3 is generated by three vectors: (1, 0, 0),(0, 1, 0), and (0, 0, 1). If the set {(1, 4,−6), (1, 5, 8), (2, 1, 1), (0, 1, 0)} was linearly independent,the replacement theorem would imply 4 ≤ 3, a contradiction.
6. Give three different bases for F2 and M2×2(F ).
Solution. For F2, we can use
{(1, 0), (0, 1)} {(−1, 1), (0, 1)} {(1,−1), (1, 0)}
For M2×2(F ), we can use{(1 00 0
),
(0 00 1
)} {(−1 10 0
),
(0 00 1
)} {(1 −10 0
),
(0 01 0
)}7. The vectors u1 = (2,−3, 1), u2 = (1, 4,−2), u3 = (−8, 12,−4), u4 = (1, 37,−17), and u5 =
(−3,−5, 8) generate R3. Find a subset of the set {u1, u2, u3, u4, u5} that is a basis for R3.
Solution. Start with the set {u1} and note that u2 is not a multiple of u1. Therefore{u1, u2} will be linearly independent. Next, guess that u5 is not a linear combination ofthe two vectors already in the set. To verify, solve the usual homogeneous system of linearequations to determine whether {u1, u2, u5} is linearly independent. As it turns out, it is, andtherefore by a corollary to the replacement theorem, the set generates R3, and so is a basisfor R3 also.
9. The vectors u1 = (1, 1, 1, 1), u2 = (0, 1, 1, 1), u3 = (0, 0, 1, 1), and u4 = (0, 0, 0, 1) form a basisfor F4. Find the unique representation of an arbitrary vector (a1, a2, a3, a4) in F4 as a linearcombination of u1, u2, u3, and u4.
22
Solution. We wish to find scalars c1, c2, c3, c4 ∈ F such that
c1 = a1
c1 + c2 = a2
c1 + c2 + c3 = a3
c1 + c2 + c3 + c4 = a4
We clearly obtain (a1, a2 − a1, a3 − a2, a4 − a3) as our representation.
10. In each part, use the Lagrange interpolation formula to construct the polynomial of smallestdegree whose graph contains the following points.
(a) (−2,−6), (−1, 5), (1, 3)
Solution. Using the notation from the book, we set c0 = −2, c1 = −1, c2 = 1, b0 = −6,b1 = 5, b2 = 3. Formulating the Lagrange polynomials, we have:
f0(x) =2∏
k=0k 6=0
(x− ck
c0 − ck
)=(
x + 1−1
)(x− 1−3
)=
13(x2 − 1)
f1(x) =2∏
k=0k 6=1
(x− ck
c1 − ck
)=(
x + 21
)(x− 1−2
)= −1
2(x2 + x− 2)
f2(x) =2∏
k=0k 6=2
(x− ck
c2 − ck
)=(
x + 23
)(x + 1
2
)=
16(x2 + 3x + 2)
We then use these polynomials in a linear combination, with the coefficient of each fi
equal to bi:
g(x) =2∑
i=0
bifi(x)
= −6[13(x2 − 1)
]+ 5
[−1
2(x2 + x− 2)
]+ 3
[16(x2 + 3x + 2)
]=(−2x2 + 2
)+(−5
2x2 − 5
2x + 5
)+(
12x2 +
32x + 1
)= −4x2 − x + 8
(c) (−2, 3), (−1,−6), (1, 0), (3,−2)
Solution. Using the notation from the book, we set c0 = −2, c1 = −1, c2 = 1, c3 = 3,
23
b0 = 3, b1 = −6, b2 = 0, b3 = −2. Formulating the Lagrange polynomials, we have:
f0(x) =3∏
k=0k 6=0
(x− ck
c0 − ck
)=(
x + 1−1
)(x− 1−3
)(x− 3−5
)= − 1
15(x3 − 3x2 − x + 3)
f1(x) =3∏
k=0k 6=1
(x− ck
c1 − ck
)=(
x + 21
)(x− 1−2
)(x− 3−4
)=
18(x3 − 2x2 − 5x + 6)
f2(x) =3∏
k=0k 6=2
(x− ck
c2 − ck
)=(
x + 23
)(x + 1
2
)(x− 3−2
)= − 1
12(x3 − 7x + 6)
f3(x) =3∏
k=0k 6=3
(x− ck
c3 − ck
)=(
x + 25
)(x + 1
4
)(x− 1
2
)=
140
(x3 + 2x2 − x− 2)
Note that there was no actual use in calculating f2(x) since b2 = 0. We then use thesepolynomials in a linear combination, with the coefficient of each fi equal to bi:
g(x) =3∑
i=0
bifi(x)
= 3[− 1
15(x3 − 3x2 − x + 3)
]− 6
[18(x3 − 2x2 − 5x + 6)
]− 2
[140
(x3 + 2x2 − x− 2)]
= −x3 + 2x2 + 4x− 5
11. Let u and v be distinct vectors of a vector space V. Show that if {u, v} is a basis for V and aand b and nonzero scalars, then both {u + v, au} and {au, bv} are also bases for V.
Solution. Since {u, v} is a basis for V, each vector w ∈ V can be written uniquely asw = a1u + a2v. Note
a2(u + v) +a1 − a2
a(au) = a2u + a2v + a1u− a2u
= a1 + a2v
= w
=a1
a(au) +
a2
b(bv)
and therefore {u + v, au} and {au, bv} generate V. Now, since {u, v} is a basis for V, theequation
a1u + a2v = 0
is satisfied only by a1 = a2 = 0. Therefore, the equation a1(u + v) + a2(au) = 0 can berewritten as (a1 + a2a)u + a1v = 0, whereby we obtain a1 = 0 and a1 + a2a = 0 by the linearindependence of {u, v}. This last equation yields a2a = 0, however a 6= 0 as stated by theexercise, which implies that a2 = 0. This proves the linear independence of {u+v, au}. Next,we examine {au, bv}, which gives the equation a1(au) + a2(bv) = 0. This can be rewritten as
24
(a1a)u + (a2b)v = 0. Again, a 6= 0 and b 6= 0, which gives a1 = a2 = 0, proving the linearindependence of {au, bv}. Therefore both sets are linearly independent and generate V, soboth are bases for V.
13. The set of solutions to the system of linear equations
x1 − 2x2 + x3 = 02x1 − 3x2 + x3 = 0
is a subspace of R3. Find a basis for this subspace.
Solution. Solving this system, we obtain that our solution set is the set
S = {(x1, x2, x3) : x1 = x2 = x3}
Therefore, a basis for this set could be {(1, 1, 1)}. The aforementioned set is linearly indepen-dent, since a1(1, 1, 1) = (0, 0, 0) clearly implies a1 = 0. It can also be readily seen that anyvector in S can be expressed in the form a(1, 1, 1) where a ∈ R.
14. Find bases for the following subspaces of F5:
W1 = {(a1, a2, a3, a4, a5) ∈ F5 : a1 − a3 − a4 = 0}
andW2 = {(a1, a2, a3, a4, a5) ∈ F5 : a2 = a3 = a4 and a1 + a5 = 0}
What are the dimensions of W1 and W2?
Solution. For W1, consider the set β1 = {(0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (1, 0, 1, 0, 0), (1, 0, 0, 1, 0)}.For W2, consider the set β2 = {(0, 1, 1, 1, 0), (1, 0, 0, 0,−1)}. Let v1 ∈ W1. Then v1 =(a1, a2, a3, a4, a5) satisfying a1 − a3 − a4 = 0, that is, a1 = a3 + a4. Now, note
a2(0, 1, 0, 0, 0) + a5(0, 0, 0, 0, 1) + a3(1, 0, 1, 0, 0) + a4(1, 0, 0, 1, 0) = (a3 + a4, a2, a3, a4, a5)= (a1, a2, a3, a4, a5)= v1
This proves that β1 spans W1. Next, let v2 ∈ W2. Then v2 = (b1, b2, b3, b4, b5) satisfyingb2 = b3 = b4 and a1 + a5 = 0. The latter equation implies a1 = −a5. Note
b2(0, 1, 1, 1, 0) + b1(1, 0, 0, 0,−1) = (b1, b2, b2, b2,−b1)= (b1, b2, b3, b4, b5)= v2
This proves that β2 spans W2. It is trivial to verify that both β1 and β2 are linearly inde-pendent. β1 and β2 therefore constitute bases for W1 and W2, respectively. Furthermore,because β1 contains 4 vectors and β2 contains 2 vectors, we conclude that W1 and W2 are ofdimension 4 and 2, respectively.
25
15. The set of all n × n matrices having trace equal to zero is a subspace W of Mn×n(F ) (seeExample 4 of Section 1.3). Find a basis for W. What is the dimension of W?
Solution. For each i, j satisfying 1 ≤ i ≤ n, 1 ≤ j ≤ n, we define Sij ∈ Mn×n(F ) in thefollowing way:
Sijkl =
{1 if k = i and l = j
0 otherwise
Next, for each i satisfying 2 ≤ i ≤ n, define Di ∈ Mn×n(F ) in the following way:
Dikl =
−1 if k = l = 1
1 if k = l = i
0 otherwise
Consider the set β = {Sij : i 6= j} ∪ {Di : 2 ≤ i ≤ n}. We will prove this set is a basis for W.First, we show β is linearly independent. We need to show that the equation
n∑i=2
aiDi +
∑1≤i≤n1≤j≤n
i 6=j
bijSij = 0
holds only when, for each i satisfying 2 ≤ i ≤ n, we have ai = 0, and for each i, j satisfying1 ≤ i ≤ n, 1 ≤ j ≤ n, and i 6= j, we have bij = 0. First, note that the matrix 0 has 0 for allof its entries. For each k satisfying 2 ≤ k ≤ n, we obtain
0 =n∑
i=2
aiDikk +
∑1≤i≤n1≤j≤n
i 6=j
bijSijkk =
n∑i=2
aiDikk +
∑1≤i≤n1≤j≤n
i 6=j
bij · 0 =n∑
i=2
aiDikk
To understand why this is, note that Sijkk is only nonzero when k = i and k = j. However
this can never be true in this case because the summation specifies i 6= j. Hence all the Sijkk
are zero. Furthermore note that for all i satisfying 2 ≤ i ≤ n, Dikk = 1 only when i = k. This
permits us to simplify the equation even more:
n∑i=2
aiDikk = ak · 1 +
∑2≤i≤n
i 6=k
ai · 0 = ak
Therefore, ak = 0 (for 2 ≤ k ≤ n). We now wish to show that bij = 0 for all 1 ≤ i ≤ n,1 ≤ j ≤ n, and i 6= j. For each k, l satisfying 1 ≤ k ≤ n, 1 ≤ l ≤ n, and k 6= l, we have
0 =n∑
i=2
aiDikl +
∑1≤i≤n1≤j≤n
i 6=j
bijSijkl =
n∑i=2
ai · 0 + bkl · 1 +∑
1≤i≤n1≤j≤n
i 6=j(∗∗)
bij · 0 = bkl
26
where the (**) denotes the condition that the equations i = k and j = l do not hold simulta-neously. To understand why this is, note that Di
kl = 0 always (since k 6= l). Note also thatSij
kl = 0 unless i = k and j = l. We therefore pull out the unique term bkl · Sklkl from the
summation. Sklkl = 1 by definition, but all the other Sij
kl = 0, and so we are left with bkl = 0(for 1 ≤ k ≤ n, 1 ≤ l ≤ n, and k 6= l). Therefore we have shown that the only representationof the zero matrix as a linear combination of the vectors within β is the trivial one, provingthat β is linearly independent.
Now we need to show that span(β) = W. Let M ∈ W and denote M ’s entry in row i, columnj by Mij . Then, since tr(M) = 0, we have
M11 = −M22 − . . .−Mnn
We will define the matrix M ′ by
M ′ =n∑
i=2
MiiDi +
∑1≤i≤n1≤j≤n
i 6=j
MijSij
and consider the entry in the upper-left corner
M ′11 =
n∑i=2
MiiDi11 +
∑1≤i≤n1≤j≤n
i 6=j
MijSij11 =
n∑i=2
Mii · (−1) +∑
1≤i≤n1≤j≤n
i 6=j
Mij · 0 =n∑
i=2
−Mii = M11
To make the above manipulations, we simply noted that Di11 = −1 (for 2 ≤ i ≤ n), and
that Sijkk = 0 (for 1 ≤ i ≤ n, 1 ≤ j ≤ n, and i 6= j). With the same kind of reasoning, for
2 ≤ k ≤ n, we obtain
M ′kk =
n∑i=2
MiiDikk +
∑1≤i≤n1≤j≤n
i 6=j
MijSijkk = Mkk · 1 +
∑2≤i≤n
i 6=k
Mii · 0 +∑
1≤i≤n1≤j≤n
i 6=j
Mij · 0 = Mkk
It now remains to verify that M ′kl = Mkl when 1 ≤ k ≤ n, 1 ≤ l ≤ n, and k 6= l. We obtain
M ′kl =
n∑i=2
MiiDikl +
∑1≤i≤n1≤j≤n
i 6=j
MijSijkl =
n∑i=2
Mii · 0 + Mkl · 1 +∑
1≤i≤n1≤j≤n
i 6=j(∗∗)
Mij · 0 = Mkl
where the (**) denotes the condition that the equations i = k and j = l do not hold si-multaneously. This completes our proof that the entries of M ′ are exactly those of M , andso M ′ = M , which demonstrates W ⊆ span(β). Clearly we also have span(β) ⊆ W, forotherwise, noting that each vector in β is also in W, W would not be closed under linearcombination, contradicting the fact that it is a subspace, which was given in the exercise.Therefore span(β) = W. This, as well as the linear independence of β, demonstrates that β
27
is indeed a basis for W, completing the proof. There are n − 1 vectors in {Di : 2 ≤ i ≤ n}and n2 − n vectors in {Sij : i 6= j}. Since β is the union of these two sets (note that thetwo sets are clearly disjoint as a result of the linear independence of β), β contains exactly(n− 1) + (n2 − n) = n2 − 1 vectors.
16. The set of all upper triangular n× n matrices is a subspace W of Mn×n(F ) (see Exercise 12of Section 1.3). Find a basis for W. What is the dimension of W?
Solution. For each i, j satisfying 1 ≤ i ≤ n, 1 ≤ j ≤ n, we define
Sijkl =
{1 if k = i and l = j
0 otherwise
and form the set β = {Sij : i ≤ j}. We will prove β is a basis for W. To verify the linearindependence of β, we have to show that the equation∑
1≤i≤n1≤j≤n
i≤j
aijSij = 0
holds only when aij = 0 for each i, j satisfying 1 ≤ i ≤ n, 1 ≤ j ≤ n, and i ≤ j. So considerthe entry of this linear combination at row k, column l, for 1 ≤ k ≤ n, 1 ≤ l ≤ n, and k ≤ l.We obtain
0 =∑
1≤i≤n1≤j≤n
i≤j
aijSij = akl · 1 +
∑1≤i≤n1≤j≤n
i≤j(∗∗)
aij · 0
and thus akl = 0 for all 1 ≤ k ≤ n, 1 ≤ l ≤ n, and k ≤ l. This proves that β is linearlyindependent. We now demonstrate that β generates W. Assume M ∈ W, that is, M is anupper triangular matrix. We define the matrix M ′ ∈ Mn×n(F ) by
M ′ =∑
1≤i≤n1≤j≤n
i≤j
MijSij
and examine the entries of this matrix
M ′kl =
∑1≤i≤n1≤j≤n
i≤j
MijSijkl = Mkl · 1 +
∑1≤i≤n1≤j≤n
i≤j(∗∗)
Mij · 0 = Mkl
where the (**) denotes the condition that the equations i = k and j = l do not hold si-multaneously. This is true because Sij
kl is defined to be 0 unless i = k and j = l. We havetherefore demonstrated that M ′ = M . This shows that W ⊆ span(β), and for the samereasons as in exercise 15, that span(β) = W. Therefore β is a basis for W. Since β containsn + (n− 1) + (n− 2) + . . . + (n− (n− 1)) = n2− (1 + 2 + . . . + (n− 1)) = 1
2 (n2 + n) elements,we conclude dim(W) = 1
2 (n2 + n).
28
17. The set of all skew-symmetric n × n matrices is a subspace W of Mn×n(F ) (see Exercise 28of Section 1.3). Find a basis for W. What is the dimension of W?
Solution. A skew-symmetric matrix is a matrix A such that At = −A. Since the diagonalsof a matrix are invariant under the matrix transpose operation, special attention must bepaid here, for in a field of characteristic 2, a = −a does not imply a = 0, and every element isits own additive inverse. The definition of skew-symmetry is equivalent to that of symmetryfor matrices with their entries in a field of characteristic 2 (−A = A in such a field, so theequation At = −A becomes At = A which is precisely the definition of a symmetric matrix).We will assume F ’s characteristic is not 2.
For each i, j satisfying 1 ≤ i ≤ n, 1 ≤ j ≤ n, we define
Sijkl =
1 if k = i and l = j
−1 if k = j and l = i
0 otherwise
and form the set β = {Sij : i < j}. The proof that β is a basis is very similar to those givenin exercises 15 and 16, so I will not repeat it. It is trivial to see that dim(W) = 1
2n(n− 1).
19. Complete the proof of Theorem 1.8.
Solution. We wish to prove that if V is a vector space and β = {u1, u2, . . . , un} is a subsetof V such that every v ∈ V can be uniquely expressed as a linear combination of vectors in β,then β is a basis for V.
Note first that β is trivially a generating set for V, since we are told each v ∈ V is also inspan(β). We know that if v = a1u1 + . . . + anun = b1u1 + . . . + bnun, then ai = bi for eachi such that 1 ≤ i ≤ n for otherwise each v would not be uniquely expressible as a linearcombination of vectors from β. This implies that ai − bi = 0 for each i in the equation,proving that β is linearly independent. β is therefore a basis for W.
20. Let V be a vector space having dimension n, and let S be a subset of V that generates V.
(a) Prove that there is a subset of S that is a basis for V. (Be careful not to assume that Sis finite.)
Solution. Since V is finite-dimensional (with dimension n), then some set β = {β1, . . . , βn}is a basis for V. Since β ⊆ V and S generates V, then for each i satisfying 1 ≤ i ≤ n,there exists some positive integer ci such that we can write
βi = ai1s
i1 + . . . + ai
cisi
ci
for some ai1, . . . , a
ici∈ F and si
1, . . . , sici∈ S. For each i such that 1 ≤ i ≤ n, define
Si = {si1, . . . , s
ici}
and let
γ =n⋃
i=1
Si
29
Clearly β ⊆ span(γ) from above, and so by Theorem 1.5 we have span(β) = V ⊆ span(γ).Clearly, span(γ) ⊆ V since V is a vector space, and therefore span(γ) = V. Theorem 1.9implies that there exists some γ′ ⊆ γ such that γ′ is a basis for V. This set γ′ is clearlya subset of S, since γ itself is a subset of S.
(b) Prove that S contains at least n vectors.
Solution. Since γ′ is a basis for V (which is of dimension n), γ′ must contain n vectors,by Corollary 1 to the Replacement Theorem. The fact that γ′ ⊆ S then gives us that Scontains at least n vectors.
1.7 Maximal Linearly Independent Subsets
Section 1.7 presents an alternative formulation of the concept of a basis. First, it defines theconcept of a chain and a maximal set (with respect to set inclusion). Then, a principle knownas the Maximal Principle (also known as the Hausdorff Maximal Principle) is presented, which isequivalent to the Axiom of Choice. Maximal linearly independent subsets are defined as sets thatare linearly independent and are contained in no linearly independent subset with the exception ofthemselves. The equivalence between maximal linearly independent subsets and bases is proved.At the end of the section, it is shown that if S is a linearly independent subset of a vector space,there exists a maximal linearly independent subset of that vector space containing S, giving theobvious but important corollary that every vector space has a basis.
4. Let W be a subspace of a (not necessarily finite-dimensional) vector space V. Prove that anybasis for W is a subest of a basis for V.
Solution. Let β be a basis for W. β is clearly a linearly independent subset of W and henceof V. Therefore by Theorem 1.12, V possesses a maximal linearly independent subset (basis)containing β.
2 Linear Transformations and Matrices
2.1 Linear Transformations, Null spaces, and Ranges
In Section 2.1, linear transformations are introduced, as a special kind of function from one vectorspace to another. Some properties of linear transformations are discussed. Two special subspaces(of the domain and codomain, respectively), the nullspace and range, are presented, and their di-mensions are called the nullity and rank, respectively. Furthermore, results are proven concerningthe injectivity and surjectivity of linear transformations satisfying certain criteria.
For Exercises 2 through 6, prove that T is a linear transformation, and find bases for both N(T)and R(T). Then compute the nullity and rank of T, and verify the dimension theorem. Finally, usethe appropriate theorems in this section to determine whether T is one-to-one or onto.
2. T : R3 → R2 defined by T(a1, a2, a3) = (a1 − a2, 2a3).
Solution. Let a = (a1, a2, a3) and b = (b1, b2, b3). Also let c ∈ R. Then T(ca + b) =(ca1+b1−ca2−b2, 2ca3+2b3) = (ca1−ca2, 2ca3)+(b1−b2, 2b3) = c(a1−a2, 2a3)+(b1−b2, 2b3) =
30
cT(a) + T(b). This proves that T is linear. Consider the set β = {(1, 1, 0)}. Let v ∈ span(β).Then v = (a, a, 0) for some a ∈ R, so that T(v) = T(a, a, 0) = (a − a, 2 · 0) = (0, 0). Thisproves v ∈ N(T) and hence span(β) ⊆ N(T). We also clearly have N(T) ⊆ span(β), andso β spans the nullspace of T. β is clearly linearly independent since it consists of a singlenonzero vector, so β is a basis for T’s nullspace. Let β′ = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. ByTheorem 2.2, R3 = span(T(β′)) = span({(1, 0), (−1, 0), (0, 2)}) = span({(1, 0), (0, 1)}). Sincethe set {(1, 0), (0, 1)} is linearly independent, it constitutes a basis for the range of T. Thus,nullity(T) = 1 and rank(T) = 2. Therefore 1 + 2 = 3, verifying the dimension theorem. Sincenullity(T) 6= 0, we see that N(T) contains some nonzero vector. By Theorem 2.4, then, Tis not one-to-one. Note that rank(T) = dim(R2), proving by Theorem 1.11 that R(T) = W.Therefore T is onto.
6. T : Mn×n(F ) → F defined by T(A) = tr(A). Recall (Example 4, Section 1.3) that
tr(A) =n∑
i=1
Aii.
Solution. Let A,B ∈ Mn×n(F ), and c ∈ F . Then T(cA+B) = tr(cA+B) = c tr(A)+tr(B) =cT(A)+T(B), by Exercise 6 of Section 1.3, proving the linearity of T. By exercise 15 of Section1.6, a basis for N(T) is
{Sij : i 6= j} ∪ {Di : 2 ≤ i ≤ n}and the nullity of T is n2 − 1, where Sij and Di are defined as in Exercise 15 of Section1.6. It is trivial to see that rank(T) = 1, and that {1} is a basis for R(T). Thereforenullity(T) + rank(T) = (n2 − 1) + 1 = n2 = dim(Mn×n(F )), verifying the dimension theorem.Since rank(T) = 1 = dim(F ), we have R(T) = F by Theorem 1.11, proving that T is onto.However, since N(T) 6= {0} when n > 2, clearly, we have by Theorem 2.4 that T is notone-to-one unless n = 1 (since in this case, nullity(T) = 12 − 1 = 0, implying N(T) = {0}).
7. Prove properties 1, 2, 3, and 4 on page 65.
Solution. Let T : V → W be linear. Clearly T(0) = 0, for otherwise by linearity T(0) =T(x+(−x)) = T(x)+T(−x) = T(x)+(−T(x)) 6= 0 which is absurd. Also note that since T islinear, T(cx+y) = T(cx)+T(y) = cT(x)+T(y), and T(x−y) = T(x+(−y)) = T(x)+T(−y) =T(x)−T(y). To prove property 4, note that if T is linear, an inductive argument can be usedto show that for all x1, . . . , xn ∈ V and a1, . . . , an ∈ F , we have
T
(n∑
i=1
aixi
)=
n∑i=1
T(aixi) =n∑
i=1
aiT(xi)
Now, assume T(cx + y) = cT(x) + T(y) for all x, y ∈ V and c ∈ F . Let x, y ∈ V. Then weobtain T(x + y) = T(1x + y) = 1 · T(x) + T(y) = T(x) + T(y). Next, let x ∈ V and c ∈ F .Then we obtain T(cx) = T(cx + 0) = c · T(x) + T(0) = c · T(x). This proves T is linear. Thesame type of reasoning can be used to show that if T satisfies
T
(n∑
i=1
aixi
)=
n∑i=1
aiT(xi)
then T must be linear.
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10. Suppose that T : R2 → R2 is linear, T(1, 0) = (1, 4), and T(1, 1) = (2, 5). What is T(2, 3)? IsT one-to-one?
Solution. By Theorem 2.6, T is defined by
T(ae1 + be2) = aw1 + bw2
So that we obtain w1 = (1, 4) and w2 = (2, 5) − (1, 4) = (1, 1). Therefore we concludeT(2, 3) = 2w1 + 3w2 = 2(1, 4) + 3(1, 1) = (2, 8) + (3, 3) = (5, 11). Now, consider x ∈ N(T).Clearly, x = ae1 + be2, so T(x) = aT(e1)+ bT(e2) = aw1 + bw2 for some scalars a, b. However,{w1, w2} is linearly independent, so that a = b = 0. This implies that N(T) = {0}, implyingT is one-to-one by Theorem 2.4.
11. Prove that there exists a linear transformation T : R2 → R3 such that T(1, 1) = (1, 0, 2) andT(2, 3) = (1,−1, 4). What is T(8, 11)?
Solution. Since {(1, 1), (2, 3)} is a basis for R2, by Theorem 2.6, there exists a unique lineartransformation such that T(1, 1) = (1, 0, 2) and T(2, 3) = (1,−1, 4). This transformation isdefined by
T(a(1, 1) + b(2, 3)) = a(1, 0, 2) + b(1,−1, 4)
To determine T(8, 11), we write (8, 11) = 2(1, 1) + 3(2, 3). This gives T(8, 11) = 2(1, 0, 2) +3(1,−1, 4) = (2, 0, 4) + (3,−3, 12) = (5,−3, 16).
12. Is there a linear transformation T : R3 → R2 such that T(1, 0, 3) = (1, 1) and T(−2, 0,−6) =(2, 1)?
Solution. There clearly exists no such linear transformation. Assume T(1, 0, 3) = (1, 1).Then by the linearity of T we must have T(−2, 0,−6) = −2·T(1, 0, 3) = −2(1, 1) = (−2,−2) 6=(2, 1).
13. Let V and W be vector spaces, let T : V → W be linear, and let {w1, w2, . . . , wk} be a linearlyindependent subset of R(T). Prove that if S = {v1, v2, . . . , vk} is chosen so that T(vi) = wi
for i = 1, 2, . . . , k, then S is linearly independent.
Solution. Assume, for the sake of contradiction, that S is linearly dependent. Then thereexist a1, . . . , ak ∈ F , not all zero, such that
a1v1 + a2v2 + . . . + akvk = 0
Then by the linearity of T, such scalars also satisfy
a1T(v1) + a2T(v2) + . . . + akT(vk) = 0
Now recall that T(vi) = wi for 1 ≤ i ≤ k. This implies that {w1, w2, . . . , wk} is a linearlydependent subset of R(T), a contradiction. Therefore S must be linearly independent.
14. Let V and W be vector spaces and T : V → W be linear.
(a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V ontolinearly independent subsets of W.
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Solution. Assume T carries linearly independent subsets of V onto linearly independentsubsets of W. Assume, for the sake of contradiction, that there exist distinct x, y ∈ Vsatisfying T(x) = T(y). Then T(x)−T(y) = 0 which implies x−y ∈ N(T) by linearity. LetS = {x−y}. Since x−y 6= 0 (for otherwise x = y), we have that S is linearly independent.However, T(S) = {0}, demonstrating the existence of a linearly independent subset of Vthat T carries onto a linearly dependent subset of W, which is impossible by assumption.Therefore no distinct x, y exist satisfying T(x) = T(y). This proves that T is one-to-one.
Assume next that T is one-to-one and that S is a linearly independent subset of V.Assume, for the sake of contradiction, that T(S) is linearly dependent. Then there existscalars a1, . . . , an, not all zero, and vectors v1, . . . , vn ∈ S such that
a1T(v1) + . . . + anT(vn) = 0
However linearity implies that
0 = T(0) = T(a1v1 + . . . + anvn)
Now, due to the linear independence of S, we see that a1v1 + . . . + anvn 6= 0 unlessall the ai are zero, which we have assumed not to be the case. This contradicts that Tis one-to-one. Therefore T(S) must be linearly independent, and so T carries linearlyindependent subsets of V onto linearly independent subsets of W.
(b) Suppose that T is one-to-one and that S is a subset of V. Prove that S is linearlyindependent if and only if T(S) is linearly independent.
Solution. Clearly, if S is linearly independent, then T(S) is linearly independent by (a).Therefore, assume T(S) is linearly independent, and for the sake of contradiction thatS is linearly dependent. Then there exist scalars a1, . . . , an, not all zero, and vectorsv1, . . . , vn ∈ S such that
a1v1 + . . . + anvn = 0
Then the same scalars will, by linearity of T, satisfy
a1T(v1) + . . . + anT(vn) = 0
This implies that T(S) is linearly dependent, a contradiction. Therefore S is linearlyindependent.
(c) Suppose β = {v1, v2, . . . , vn} is a basis for V and T is one-to-one and onto. Prove thatT(β) = {T(v1),T(v2), . . . ,T(vn)} is a basis for W.
Solution. Since T is one-to-one, (b) gives us that T(β) is linearly independent. Since βis a basis for V, Theorem 2.2 gives us that R(T) is generated by β. However, R(T) = Wsince T is onto. Therefore β is a basis for W.
17. Let V and W be finite-dimensional vector spaces and T : V → W be linear.
(a) Prove that if dim(V) < dim(W), then T cannot be onto.
Lemma. Let V and W are vector spaces, T : V → W be linear, and S ⊆ V. If T(S) islinearly independent, then S is linearly independent.
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Proof. Assume T(S) is linearly independent, and for the sake of contradiction, thatS is linearly dependent. Then there exist scalars a1, . . . , an, not all zero, and vectorss1, . . . , sn ∈ S such that
a1s1 + . . . + ansn = 0
However, T(0) = 0, so by linearity we would obtain that T(S) is linearly dependent, acontradiction. Therefore S is linearly independent.
Solution. Let dim(V) < dim(W) and assume, for the sake of contradiction, that T isonto. Let β be a basis for W. Then β contains dim(W) elements. Since T is onto, β= T(S) for some subset S ⊆ V. However, β is clearly linearly independent in W, soby the lemma we see S is linearly independent in V. Therefore V contains a linearlyindependent subset with dim(W) elements. Since dim(W) > dim(V), this contradictsthe replacement theorem. Therefore T cannot be onto.
(b) Prove that if dim(V) > dim(W), then T cannot be one-to-one.
Solution. If T is one-to-one, then T carries linearly independent subsets of V ontolinearly independent subsets of W. Let β be a basis for V. Then β contains dim(V)elements, and is linearly independent. So T(β) will be a linearly independent subset ofW containing dim(V) elements, by (b) of Exercise 14. However, dim(W) < dim(V), sowe obtain the same contradiction to the replacement theorem as in (a). Thus, T cannotbe one-to-one.
2.2 The Matrix Representation of a Linear Transformation
Section 2.2 discusses a correspondence between linear transformations and matrices. It defines thematrix representation of a linear transformation in such a way that the columns of said matrix canbe deduced simply by examining the image of each vector in the domain’s basis, represented as acoordinate vector with respect to the basis of the codomain. The vector space of linear transforma-tions from V to W, L(V,W), is also defined. We also see that sums and scalar multiples of lineartransformations have matrix representations as we would expect, i.e. the matrix representation of asum of transformations is the sum of the matrix representations of the individual transformations.
2. Let β and γ be the standard ordered bases for Rn and Rm, respectively. For each lineartransformation T : Rn → Rm, compute [T]γβ .
(a) T : R2 → R3 defined by T(a1, a2) = (2a1 − a2, 3a1 + 4a2, a1).
Solution. Recall that the jth column of [T]γβ is the column matrix [T(βj)]γ . Sincewe’re working with the standard ordered bases here, this simplifies to [T(ej)]γ . NoticeT(e1) = T(1, 0) = (2 − 0, 3 + 0, 1) = (2, 3, 1) = 2e1 + 3e2 + 1e3, and T(e2) = T(0, 1) =(0− 1, 0 + 4, 0) = (−1, 4, 0) = −1e1 + 4e2 + 0e3. Therefore
[T]γβ =
2 −13 41 0
(c) T : R3 → R defined by T(a1, a2, a3) = 2a1 + a2 − 3a3.
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Solution. Applying the same method as before, we see T(1, 0, 0) = 2 = 2 ·1, T(0, 1, 0) =1 = 1 · 1, and T(0, 0, 1) = −3 = −3 · 1. Therefore
[T]γβ =(2 1 −3
)(e) T : Rn → Rn defined by T(a1, a2, . . . , an) = (a1, a1, . . . , a1).
Solution. Let A be the n× n matrix defined by
Aij =
{1 if j = 10 otherwise
Then [T]γβ = A.
3. Let T : R2 → R3 be defined by T(a1, a2) = (a1 − a2, a1, 2a1 + a2). Let β be the standardordered basis for R2 and γ = {(1, 1, 0), (0, 1, 1), (2, 2, 3)}. Compute [T]γβ . If α = {(1, 2), (2, 3)},compute [T]γα.
Solution. We see that T(1, 0) = (1, 1, 2) = − 13 (1, 1, 0) + 0(0, 1, 1) + 2
3 (2, 2, 3), and T(0, 1) =(−1, 0, 1) = −1(1, 1, 0)+1(0, 1, 1)+0(2, 2, 3). Therefore, the coordinate vectors for T(e1) andT(e2) are (− 1
3 , 0, 23 ) and (−1, 1, 0) respectively. So we obtain
[T]γβ =
− 13 −10 123 0
Next, we see that T(1, 2) = (−1, 1, 4) = − 7
3 (1, 1, 0) + 2(0, 1, 1) + 23 (2, 2, 3) and T(2, 3) =
(−1, 2, 7) = − 113 (1, 1, 0) + 3(0, 1, 1) + 4
3 (2, 2, 3). Therefore, the coordinate vectors for T(1, 2)and T(2, 3) are (− 7
3 , 2, 23 ) and (− 11
3 , 3, 43 ) respectively. So we obtain
[T]γα =
− 73 − 11
32 323
43
Note that in both cases, a system of linear equations had to be solved to express each imagevector as a combination of the vectors in the codomain’s basis, although this was not shown.
8. Let V be an n-dimensional vector space with an ordered basis β. Define T : V → Fn byT(x) = [x]β . Prove that T is linear.
Solution. Let x, y ∈ V and c ∈ F . Assume that β = {v1, . . . , vn} and that
x =n∑
i=1
aivi and y =n∑
i=1
bivi
for scalars a1, . . . , an, b1, . . . , bn. Then we have
T(cx + y) = [cx + y]β =
ca1 + b1
ca2 + b2
...can + bn
= c
a1
a2
...an
+
b1
b2
...bn
= c[x]β + [y]β = cT(x) + T(y)
which proves T is linear.
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2.3 Composition of Linear Transformations and Matrix Multiplication
Section 2.3 uses composition of linear transformations to motivate the usual definition of matrixmultiplication. Some useful theorems are proved about compositions of transformations and matrixmultiplication. An important relationship between matrices and linear transformations is demon-strated through the so-called “left-multiplication transformation”, LA.
13. Let A and B be n× n matrices. Recall that the trace of A is defined by
tr(A) =n∑
i=1
Aii
Prove that tr(AB) = tr(BA) and tr(A) = tr(At).
Solution. Note that we have
tr(BA) =n∑
i=1
(BA)ii =n∑
i=1
(n∑
k=1
BikAki
)=
n∑k=1
(n∑
i=1
AkiBik
)=
n∑k=1
(AB)kk = tr(AB)
which proves the first proposition. To prove the second, simply note
tr(At) =n∑
i=1
(At)ii =n∑
i=1
Aii = tr(A).
15. Let M and A be matrices for which the product matrix MA is defined. If the jth column of Ais a linear combination of a set of columns of A, prove that the jth column of MA is a linearcombination of the corresponding columns of MA with the same corresponding coefficients.
Solution. Assume M is an m×n matrix, and A is an n×p matrix. Then for each k satisfying1 ≤ k ≤ p and k 6= j, there exist ak, with
Aij =p∑
k=1k 6=j
akAik
for all i satisfying 1 ≤ i ≤ n. Notice that
(MA)ij =n∑
z=1
MizAzj =n∑
z=1
Miz
p∑k=1k 6=j
akAzk
=p∑
k=1k 6=j
ak
(n∑
z=1
MizAzk
)=
p∑k=1k 6=j
ak(MA)ik
which proves the proposition.
18. Using only the definition of matrix multiplication, prove that multiplication of matrices isassociative.
Solution. Assume A is an m×n matrix, B is an n×p matrix, and C is a p× q matrix. Thenwe have
[(AB)C]ij =p∑
z=1
(AB)izCzj =p∑
z=1
(n∑
y=1
AiyByz
)Czj =
n∑y=1
Aiy
(p∑
z=1
ByzCzj
)= [A(BC)]ij .
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