definitions 061108
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Denitions of Linear Algebra Terms
In order to learn and understand mathematics, it is necessary to understandthe meanings of the terms (vocabulary words) that are used. This documentcontains denitions of some of the important terms used in linear algebra.All of these denitions should be memorized (and not just memorized butunderstood). In addition to the denitions that are given here, we also includestatements of some of the most important results (theorems) of linear algebra.The proofs of these theorems are not included here but will be given in class.
There will be a question on each of the exams that asks you to statesome denitions. It will be required that you state these denitions veryprecisely in order to receive credit for the exam question. (Seemingly very
small things such as writing the word or instead of the word and, forexample, can render a denition incorrect.) The denitions given here areorganized according to the section number in which they appear in DavidLays textbook.
Here is a suggestion for how to study linear algebra (and other subjectsin higher level mathematics):
1. Carefully read a section in the textbook and also read your class notes.Pay attention to denitions of terms and examples and try to under-stand each concept along the way as you read it. You dont have tomemorize all of the denitions upon rst reading. You will memorize
them gradually as you work the homework problems.
2. Work on the homework problems. As you work on the homework prob-lems, you should nd that you will need to continuously refer back tothe denitions and to the examples given in the textbook. (This ishow you will naturally come to memorize the denitions.) You willnd that you have to spend a considerable amount of time working onhomework. When you are done, you will probably have read throughthe section and your course notes several times and you will probablyhave memorized the denitions. (There are some very good TrueFalsequestions among the homework problems that are good tests of your
understanding of the basic terms and concepts. Such TrueFalse ques-tions will also appear on the exams given in this course.)
3. Test yourself. After you are satised that you have a good under-standing of the concepts and that can apply those concepts in solving
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problems, try to write out the denitions without referring to your book
or notes. You might then also try to work some of the supplementaryproblems that appear at the end of each chapter of the textbook.
4. One other tip is that it is a good idea to read a section in the textbookand perhaps even get started on working on the homework problemsbefore we go over that section in class. This will better prepare you totake notes and ask questions when we do go over the material in class.It will alert you to certain aspects of the material that you might notquite understand upon rst reading and you can then ask questionsabout those particular aspects during class.
1.1: Systems of Linear Equations
Denition 1 A linear equation in the variables x1; x2; : : : ; xn is an equa-tion of the form
a1x1 + a2x2 + : : : + anxn = b
where a1; a2; : : : ; an and b are real or complex numbers (that are often givenin advance).
Example 24x + 4y
7z + 8w = 8
is a linear equation in the variables x;y;z; and w.
3x + 5y2 = 6
is not a linear equation (because of the appearance of y2).
Denition 3 Suppose that m and n are positive integers. A system oflinear equations in the variables x1; x2; : : : ; xn is a set of equations of the
form
a11x1 + a12x2 + : : : + a1nxn = b1
a21x1 + a22x2 + : : : + a2nxn = b2...
am1x1 + am2x2 + : : : + amnxn = bm
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whereaij (i = 1; : : : ; m, j = 1; : : : n) andbi (i = 1; : : : ; m) are real or complex
numbers (that are often given in advance). This system of linear equationsis said to consist of m equations with n unknowns.
Example 4
2x1 + 2x2 5x3 + 4x4 2x5 = 44x1 + x2 5x4 = 1
3x1 2x2 + 4x4 2x5 = 4is a system of three linear equations in ve unknowns. (The unknowns orvariables of the system are x1; x2; x3; x4; and x5.)
Denition 5 A solution of the linear system
a11x1 + a12x2 + : : : + a1nxn = b1
a21x1 + a22x2 + : : : + a2nxn = b2...
am1x1 + am2x2 + : : : + amnxn = bm
is an orderedntuple of numbers (s1; s2; : : : ; sn) such that all of the equationsin the system are true when the substitution x1 = s1; x2 = s2; : : : ; xn = sn ismade.
Example 6 The5tuple 0;1;12
5
; 0; 3 is a solution of the linear system2x1 + 2x2 5x3 + 4x4 2x5 = 4
4x1 + x2 5x4 = 13x1 2x2 + 4x4 2x5 = 4
because all equations in the system are true when we make the substitutions
x1 = 0
x2 = 1x3 = 12
5
x4 = 0x5 = 3.
The 5tuple
1; 3; 35
; 0; 12
is also a solution of the system (as the reader can
check).
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Denition 7 Thesolution set of a system of linear equations is the set of
all solutions of the system.
Denition 8 Two linear systems are said to be equivalent to each other ifthey have the same solution set.
Denition 9 A linear system is said to be consistent if it has at least onesolution and is otherwise said to be inconsistent.
Example 10 The linear system
2x1 + 2x2 5x3 + 4x4 2x5 = 4
4x1 + x2 5x4 = 13x1 2x2 + 4x4 2x5 = 4is consistent because it has at least one solution. (It in fact has innitelymany solutions.) However, the linear system
x + 2y = 5
x + 2y = 9
is obviously inconsistent.
Denition 11 Suppose thatm andn are positive integers. Anm
n matrixis a rectangular array of numbers with m rows and n columns. Matrices areusually denoted by capital letters. If m = 1 or n = 1, then the matrix isreferred to as a vector. In particular, if m = 1, then the matrix is a rowvector of dimension n, and if n = 1 then the matrix is a column vector ofdimension m. Vectors are usually denoted by lower case boldface letters or(when handwritten) as lower case letters with arrows above them.
Example 12
A =
2
66664
8 10 6 85 7 6 60 5 10 11 3 10 54 8 1 10
3
77775
is a 5 4 matrix.r = !r = 2 3 1 9 8
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is a row vector of dimension 5.
c = !c =24 45
10
35
is a column vector of dimension 3.
Remark 13 If A is an m n matrix whose columns are made up of thecolumn vectors c1; c2; : : : ;cn, then A can be written as
A = [c1 c2 cn] .
Likewise, if A is made up of the row vectors r1; r2; : : : ; rm, then A can bewritten as
A =
26664
r1r2...rm
37775 .
Example 14 The5 4 matrix
A =266664
8 10 6 8
5 7 6
6
0 5 10 11 3 10 54 8 1 10
377775
can be written asA = [c1 c2 c3 c4]
where
c1 =
2
66664
85014
3
77775 , c2 =
2
66664
107
538
3
77775 , c3 =
2
66664
66
10101
3
77775 , c4 =
2
66664
861510
3
77775
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or can be written as
A =
266664
r1r2r3r4r5
377775
where
r1 =
8 10 6 8 r2 =
5 7 6 6 r3 =
0 5 10 1
r4 =
1 3 10 5 r5 =
4 8 1 10 .Denition 15 For a given linear system of equations
a11x1 + a12x2 + : : : + a1nxn = b1
a21x1 + a22x2 + : : : + a2nxn = b2...
am1x1 + am2x2 + : : : + amnxn = bm,
the m n matrix
A =
26664
a11 a12 a1na21 a22 a2n
......
. . ....
am1 am2 amn
37775
is called the coecient matrix of the system and the m (n + 1) matrix
[A b] =
2
6664
a11 a12 a1n b1a21 a22 a2n b2
..
.
..
.
. ..
..
.
..
.am1 am2 amn bm
3
7775
is called the augmented matrix of the system.
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Example 16 The coecient matrix of the linear system
2x1 + 2x2 5x3 + 4x4 2x5 = 44x1 + x2 5x4 = 1
3x1 2x2 + 4x4 2x5 = 4is
A =
24 2 2 5 4 24 1 0 5 0
3 2 0 4 2
35
and the augmented matrix of this system is
[A b] =24 2 2 5 4 2 44 1 0 5 0 1
3 2 0 4 2 4
35 .
Denition 17 An elementary row operation is performed on a matrixby changing the matrix in one of the following ways:
1. Interchange: Interchange two rows of the matrix. If rows i andj areinterchanged, then we use the notation ri ! rj.
2. Scaling: Multiply all entries in a single row by a nonzero constant,k. If theith row is multiplied by k, then we use the notation k
ri
!ri.
3. Replacement: Replace a row with the sum of itself and a nonzeromultiple of another row. If the ith row is replaced with the sum of itselfand k times the jth row, then we use the notationri + k rj ! ri.
Example 18 Here are some examples of elementary row operations per-formed on the matrix
A =
24 2 5 7 77 4 8 2
8 7 8 2
35 .
1. Interchange rows 1 and 3:24 2 5 7 77 4 8 2
8 7 8 2
35 r1!r3!
24 8 7 8 27 4 8 22 5 7 7
35
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2. Scale row 1 by a factor of5.24 2 5 7 77 4 8 2
8 7 8 2
35 5r1!r1!
24 10 25 35 357 4 8 2
8 7 8 2
35
3. Replace row 2 with the sum of itself and 2 times row 3.24 2 5 7 77 4 8 2
8 7 8 2
35 r2+2r3!r2!
24 2 5 7 79 18 8 2
8 7 8 2
35
Denition 19 Anm
n matrix, A, is said to berow equivalent (or simply
equivalent) to an m n matrix, B, if A can be transformed into B by aseries of elementary row operations. If A is equivalent to B, then we writeA B.Example 20 As can be seen by studying the previous example2
4 2 5 7 77 4 8 28 7 8 2
35
24 8 7 8 27 4 8 22 5 7 7
35 ,
242 5 7 7
7 4 8 28 7 8 23524
10 25 35 35
7 4 8 28 7 8 235
,
and 24 2 5 7 77 4 8 2
8 7 8 2
35
24 2 5 7 79 18 8 2
8 7 8 2
35 .
Example 21 The sequence of row operations2
43 61 6
7
1
3
52r1!r1!
2
46 121 6
7
1
3
54r3!r3!
2
46 121 628 4
3
5r1!r3!
2
428 41 6
6
12
3
5shows that 2
4 3 61 67 1
35
24 28 41 66 12
35 .
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Remark 22 (optional) Row equivalence of matrices is an example of what
is called an equivalence relation. An equivalence relation on a nonemptyset of objects, X, is a relation, , that is reexive, symmetric, and transitive.The reexive property is the property that x x for all x 2 X. The symmetric property is the property that if x 2 X and y 2 X with x y,then y x. The transitive property is the property that if x 2 X, y 2 X,and z 2 X with x y and y z, then x z. If we takeX to be the setof all m n matrices, then row equivalence is an equivalence relation on Xbecause
1. A A for all A 2 X.
2. If A 2 X and B 2 X and A B, then B A.3. If A 2 X, B 2 X, and C2 X and A B and B C, then A C.
1.2: Row Reduction and Echelon Forms
Denition 23 The leading entry in a row of a matrix is the rst (fromthe left) nonzero entry in that row.
Example 24 For the matrix
2666640 1 4 80 0 4 35 0 0 100 0 0 19 7 2 8
377775 ,
the leading entry in row 1 is 1, the leading entry in row 2 is4, the leadingentry in row 3 is 5, the leading entry in row 4 is1, and the leading entryin row 5 is 9. (Note that if a matrix happens to have a row consisting of allzeros, then that row does not have a leading entry.)
Denition 25 A zero row of a matrix is a row consisting entirely of zeros.A nonzero row is a row that is not a zero row.
Denition 26 Anechelon form matrix is a matrix for which
1. No zero row has a nonzero row below it.
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2. Each leading entry is in a column to the right of the leading entry in
the row above it.3. All entries in the same column and below any leading entry are zeros.
Denition 27 A reduced echelon form matrixis an echelon form matrixwhich also has the properties:
4. The leading entry in each nonzero row is 1.
5. Each leading 1 is the only nonzero entry in its column.
Example 28 The matrix
26646 4 101 2 104 4 20 10 1
3775
is not an echelon form matrix.The matrix
0 0 0 4 80 0 0 0 3
is an echelon form matrix but not a reduced echelon form matrix.The matrix 2
666640 0 1 00 0 0 10 0 0 00 0 0 00 0 0 0
377775
is a reduced echelon form matrix.
Theorem 29 (Uniqueness of the Reduced Echelon Form) Each ma-trix, A, is equivalent to one and only one reduced echelon form matrix. Thisunique matrix is denoted by rref(A).
Example 30 For
A =
26646 4 101 2 104 4 20 10 1
3775 ;
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we can see (after some work) that
rref(A) =
2664
1 0 00 1 00 0 10 0 0
3775 .
For
A =
0 0 0 4 80 0 0 0 3
,
we have
rref(A) = 0 0 0 1 0
0 0 0 0 1 .
For
A =
266664
0 0 1 00 0 0 10 0 0 00 0 0 00 0 0 0
377775 ,
we have
rref(A) =
2
66664
0 0 1 00 0 0 10 0 0 00 0 0 00 0 0 0
3
77775 .
Denition 31 A pivot position in a matrix, A, is a position in A thatcorresponds to a rowleading 1 in rref(A). A pivot column of A is columnof A that contains a pivot position.
Example 32 By studying the preceding example, we can see that the pivotcolumns of
A =
26646 4 101 2
10
4 4 20 10 1
3775
are columns 1, 2, and 3. (Thus all columns of A are pivot columns).
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The pivot columns of
A =
0 0 0 4 80 0 0 0 3
are columns 4 and 5.The pivot columns of
A =
266664
0 0 1 00 0 0 10 0 0 00 0 0 00 0 0 0
377775
are columns 3 and 4.
Denition 33 Let A be the coecient matrix of the linear system of equa-tions
a11x1 + a12x2 + : : : + a1nxn = b1
a21x1 + a22x2 + : : : + a2nxn = b2...
am1x1 + am2x2 + : : : + amnxn = bm.
Each variable that corresponds to a pivot column of A is called abasic vari-
able and the other variables are called free variables.
Example 34 The coecient matrix of the linear system
2x1 + 2x2 5x3 + 4x4 2x5 = 44x1 + x2 5x4 = 1
3x1 2x2 + 4x4 2x5 = 4is
A =
24
2 2 5 4 2
4 1 0
5 0
3 2 0 4 2
35 .
Since
rref(A) =
24 1 0 0
6
5
2
5
0 1 0 15
8
5
0 0 1 25
6
5
35 ,
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we see that the pivot columns of A are columns 1, 2, and 3. Thus the basic
variables of this linear system are x1, x2; and x3. The free variables are x4and x5.
Theorem 35 (Existence and Uniqueness Theorem) A linear system ofequations is consistent if and only if the rightmost column of the augmentedmatrix for the system is not a pivot column. If a linear system is consistent,then the solution set of the linear system contains either a unique solution(when there are no free variables), or innitely many solutions (whenthere is at least one free variable).
Example 36 The augmented matrix of the system
2x1 + 2x2 5x3 + 4x4 2x5 = 44x1 + x2 5x4 = 1
3x1 2x2 + 4x4 2x5 = 4
is
[A b] =
24 2 2 5 4 2 44 1 0 5 0 1
3 2 0 4 2 4
35 .
Since
rref([A b]) =24 1 0 0
6
5
2
5
6
5
0 1 0 15
85
195
0 0 1 25
6
5
6
5
35 ,we see that the rightmost column of [A b] is not a pivot column. This meansthat the system is consistent. Furthermore, since the system does have freevariables, there are innitely many solutions to the system.
Here is a general procedure for determining whether a given linear sys-tem of equations is consistent or not and, if it is consistent, whether it hasinnitely many solutions or a unique solution:
1. Compute rref([A b]) to determine whether or not the rightmost col-umn of [A b] is a pivot column. If the rightmost of [A b] is a pivotcolumn, then the system is inconsistent.
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2. Assuming that the rightmost column of [A b] is not a pivot column,
take a look at rref(A). (This requires no extra computation.) If allcolumns of A are pivot columns, then the linear system has a uniquesolution. If not all columns of A are pivot columns, then the systemhas innitely many solutions.
1.3: Vector Equations
Denition 37 Suppose that
u =26664
u1u2...
un
37775 and v = 26664v1v2...
vn
37775
are column vectors of dimension n. We dene the sumofu andv to be thevector
u + v =
26664
u1 + v1u2 + v2
...un + vn
37775 .
Denition 38 Suppose that
u =
26664
u1u2...
un
37775
is a column vectors of dimension n and suppose that c is a real number (alsocalled a scalar). We dene the scalar multiple cu to be
cu =
26664cu1cu2
...cun
37775 .
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Denition 39 The symbolRn denotes the set of all orderedntuples of num-
bers. The symbol Vn denotes the set of all column vectors of dimension n.(The textbook author does not make this distinction. He uses Rn to denoteboth things. However, there really is a dierence. Elements of Rn shouldbe regarded as points whereas elements of Vn are matrices with one column.)If x = (x1; x2; : : : ; xn) is a point in R
n, then the position vector of x is thevector
x =
26664
x1x2...
xn
37775 .
Example 40 x = (1; 2) is a point in R2
. The position vector of this point(which is an element of V2) is
x =
12
.
Denition 41 Suppose that u1;u2; : : : ;un are vectors in Vm. (Note thatthis is a set of n vectors, each of which has dimension m). Also suppose thatc1; c2; : : : ; cn are scalars. Then the vector
c1u1 + c2u2 + + cnunis called a linear combination of the vectorsu1;u2; : : : ;un.
Example 42 Suppose that
u1 =
12
, u2 =
05
, u3 =
14
.
Then the vector
4u1 2u2 + 3u3 = 4
12
2
05
+ 3
14
=
7
30
is a linear combination ofu1,u2, andu3.
Denition 43 Suppose thatu1;u2; : : : ;un are vectors in Vm. The span of
this set of vectors, denoted by Spanfu1;u2; : : : ;ung, is the set of all vectorsthat are linear combinations of the vectorsu1;u2; : : : ;un. Formally,
Spanfu1;u2; : : : ;ung= fv 2 Vm j there exist scalars c1; c2; : : : ; cn such thatv = c1u1 + c2u2 + + cnung .
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Denition 44 Thezero vector in Vm is the vector
0m =
26664
00...0
37775 .
Remark 45 Ifu1;u2; : : : ;un isanyset of vectors inVm, then0m 2 Spanfu1;u2; : : : ;ungbecause
0m = 0u1 + 0u2 + + 0un.
Denition 46 A vector equation in the variablesx1, x2, . . . ,xn is an equa-
tion of the formx1a1 + x2a2 + + xnan = b
wherea1,a2, . . . ,an andb are vectors in Vm (which are usually known before-hand). An ordered ntuple (s1; s2; : : : ; sn) is called a solution of this vectorequation if the equation is satised when we setx1 = s1, x2 = s2, . . . , xn = sn.The solution set of the vector equation is the set of all of its solutions.
Example 47 An example of a vector equation is
x1 1
2 + x2
0
5 + x3
1
4 =
42
.
The ordered triple (x1; x2; x3) = (9; 4=5; 5) is a solution of this vector equa-tion because
9
12
+
4
5
05
+ 5
14
=
42
.
Remark 48 The vector equation
x1a1 + x2a2 + + xnan = b
is consistent if and only ifb
2Span
fa1; a2; : : : ;an
g.
Remark 49 The vector equation
x1a1 + x2a2 + + xnan = b
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where
a1 =
26664a11a21
...am1
37775 , a2 =26664
a12a22
...am2
37775 , . . . ,an =26664
a1na2n
...amn
37775and
b =
26664
b1b2...
bm
37775
is equivalent to the system of linear equations
a11x1 + a12x2 + : : : + a1nxn = b1
a21x1 + a22x2 + : : : + a2nxn = b2...
am1x1 + am2x2 + : : : + amnxn = bm.
This means that the vector equation and the system have the same solutionsets. Each can be solved (or determined to be inconsistent) by rowreducingthe augmented matrix
a1 a2 an b .Example 50 Let us solve the vector equation
x1
12
+ x2
05
+ x3
14
=
42
.
This equation is equivalent to the linear system
x1 + x3 = 42x1 5x2 + 4x3 = 2.
Since
[A b] =
1 0 1 42 5 4 2
and
rref([A b]) =
1 0 1 40 1 2
56
5
,
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we see that the vector equation is consistent (because the rightmost column of
[A b] is not a pivot column) and that it in fact has innitely many solutions(because not all columns of A are pivot columns). We also see that the generalsolution is
x1 = 4 x3x2 = 6
5+
2
5x3
x3 = free.
Note that if we set x3 = 5, then we get the particular solution
x1 = 9x2 =
4
5x3 = 5
that was identied in an earlier example.
1.4: The Matrix Equation Ax = b
Denition 51 For a matrix of size m n,
A =a1 a2 an
=
26664
a11 a12 a1na21 a22 a2n
......
. . ....
am1 am2 amn
37775 ,
and a vector of dimension n,
x =
26664
x1x2...
xn
37775
,
we dene the product Ax to be
Ax = x1a1 + x2a2 + + xnan.
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Example 52 Suppose
A =
2 3 44 3 3
and
x =
24 41
0
35 .
Then
Ax = 4
24
1
33
+ 0
43
=
1119
.
Denition 53 Suppose that
A =a1 a2 an
=
26664
a11 a12 a1na21 a22 a2n
......
. . ....
am1 am2 amn
37775
is an m n matrix and that
b =
26664
b1b2
...bm
37775 2 Vm.
ThenAx = b
is called a matrix equation. Generally, A andb are known beforehand andthe problem is to solve for the unknown vectorx 2 Vn.
Remark 54 The system of linear equations
a11x1 + a12x2 + : : : + a1nxn = b1a21x1 + a22x2 + : : : + a2nxn = b2
...
am1x1 + am2x2 + : : : + amnxn = bm,
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the vector equation
x1a1 + x2a2 + + xnan = b,and the matrix equation
Ax = b
are all equivalent to each other! The only subtle dierence is that solutionsof the system and of the vector equation are ordered ntuples
(x1; x2; : : : ; xn) = (s1; s2; : : : ; sn) 2 Rn;whereas solutions of the matrix equation are the corresponding vectors
26664x1
x2...xn
37775 =26664
s1
s2...sn
37775 2 Vn.
1.5: Solution Sets of Linear Systems
Denition 55 A homogeneous linear system is one of the form
Ax = 0m
where A is an m n matrix.A nonhomogeneous linear system is a linear system that is not ho-mogeneous.
Remark 56 Homogenous systems are consistent! Thats because they havethe solutionx = 0n.
Denition 57 The solution x = 0n is called the trivial solution of thehomogeneous systemAx = 0m. A nontrivial solutionof the homogeneoussystem Ax = 0m is a solutionx 2 Vn such thatx 6= 0n.Example 58 The system
5x + 2y
3z + w = 0
3x + y 5z + w = 0x 3y 5z + 5w = 0
is homogeneous. A nontrivial solution of this system (as the reader cancheck) is (x;y;z;w) = (18; 34; 1; 25).
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Theorem 59 If every column of A is a pivot column, then the homogeneous
system Ax = 0m has only the trivial solution. If not every column of A isa pivot column, then the homogeneous system Ax = 0m has innitely manynontrivial solutions (in addition to the trivial solution).
Example 60 The system given in the preceding example has coecient ma-trix
A =
24 5 2 3 13 1 5 1
1 3 5 5
35 .
Without doing any computations, we can see that it is not possible that everycolumn of A can be a pivot column. (Why?) Therefore the homogeneous
system Ax = 0m has innitely many nontrivial solutions.
Theorem 61 Suppose that A is an m n matrix and suppose thatb 2 Vm.Also suppose thatv 2 Vn is a solution of the homogeneous system Ax = 0mand thatw 2 Vn is a solution of the system Ax = b. Thenv + w is also asolution of Ax = b.
1.7: Linear Independence
Denition 62 A set of vectors
fv1;v2; : : : ;vn
ginVm is said to be linearly
independent if the vector equation
x1v1 + x2v2 + + xnvn = 0mhas only the trivial solution. Otherwise the set of vectors is said to be linearlydependent.
Example 63 Letv1 andv2 be the vectors
v1 =
2
4414
3
5 , v2 =2
4114
3
5 .
We solve the vector equation
x1v1 + x2v2 = 03
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by observing that
A =24 4 11 14 4
3524 1 00 10 0
35 .Since each column of A is a pivot column, the vector equation has only thetrivial solution. Therefore the set of vectorsfv1;v2g is linearly independent.Denition 64 Suppose that fv1;v2; : : : ;vng is a linearly dependent set ofvectors and suppose that (c1; c2; : : : ; cn) is a nontrivial solution of the vectorequation
x1v1 + x2v2 + + xnvn = 0m.Then the equation
c1v1 + c2v2 + + cnvn = 0mis called alinear dependence relationfor the set of vectorsfv1;v2; : : : ;vng.Example 65 Consider the set of vectors
v1 =
24 414
35 , v2 =
24 114
35 , v3 =
24 23
4
35 , v4 =
24 433
35 .
We solve the vector equation
x1v1 + x2v2 + x3v3 + x4v4 = 03
by observing that24 4 1 2 41 1 3 34 4 4 3
35
24 1 0 0
11
40
0 1 0 1320
0 0 1 98
35
and hence that the vector equation has nontrivial solutions and that thegeneral solution is
x1 = 11tx2 = 26t
x3 = 45tx4 = 40t.
A linear dependence relation for the set of vectors fv1,v2,v3,v4g is11v1 + 26v2 + 45v3 + 40v4 = 03.
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1.8: Introduction to Linear Transformations
Denition 66 A transformation (orfunction ormapping), T, fromVnto Vm is a rule that assigns each vectorx 2 Vn to a vector T(x) 2 Vm. ThesetVn is called the domain of T and the set Vm is called the codomain (ortarget set) of T. If T is a transformation from Vn to Vm, then we writeT : Vn ! Vm.Denition 67 Suppose that T : Vn ! Vm. For each x 2 Vn, the vectorT (x) 2 Vm is called the image of x under the action of T. The set of allimages of vectors in Vn under the action of T is called the range of T:In setnotation, the range of T is
fy 2 Vm j y = T (x) for at least onex 2 Vng .Example 68 This example illustrates the terms dened in the preceding twodenitions: Consider the transformation T : V2 ! V3 dened by
T
xy
=
24 x
2
4y x
35 .
The domain of T is V2 and the codomain of T is V3.The image of the vector
x = 41
is
T(x) =
24 1645
35 .
The range of T consists of all vectors in V3 that have the form24 s4
t
35
where s can be any nonnegative real number and t can be any real number.For example, the vector
y =
24 649
35
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is in the range of T because
y = T
p69 + p6
.
Note that it is also true that
y = T
p69p6
.
Are there any other vectorsx 2 V2 such that T(x) = y? The answer to thisdoes not matter as far as determining whether or not y is in the range ofT is concerned. We know thaty is in the range of T because we know that
there exists at least one vectorx 2 V2 such that T(x) = y.
Denition 69 A transformationT : Vn ! Vm is said to be alinear trans-formation if both of the following conditions are satised:
1. T(u + v) = T (u) + T(v) for all vectorsu andv in Vn.
2. T(cu) = cT(u) for all vectors u in Vn and for all scalars c.
Example 70 Let T : V3 ! V2 be dened by
T
0@24 xyz
351A =
xx + y + 2z
.
We will show that T is a linear transformation: Let
u =
24 x1y1
z1
35 and v =
24 x2y2
z2
35
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be two vectors in V3. Then
T(u + v) = T
0@24 x1 + x2y1 + y2
z1 + z2
351A
=
x1 + x2
(x1 + x2) + (y1 + y2) + 2 (z1 + z2)
=
x1 + x2
(x1 + y1 + 2z1) + (x2 + y2 + 2z2)
=
x1
x1 + y1 + 2z1
+
x2
x2 + y2 + 2z2
= T(u) + T(v) :
Now let
u =
24 xy
z
35
be a vector in V3 and let c be a scalar. Then
T(cu) = T
0@24 cxcy
cz
351A
=
cxcx + cy + 2 (cz)
= c
x
x + y + 2z
cT(u) .
We have shown that T is a linear transformation.
Denition 71 The zero transformation is the linear transformation T :Vn ! Vm dened by T(u) = 0m for allu 2 Vn.
Denition 72 The identity transformation is the linear transformationT : Vn ! Vn dened by T(u) = u for allu 2 Vn.
We leave it to the reader to prove that the zero transformation and theidentity transformation are both linear transformations.
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Example 73 The transformation T : V2 ! V3 dened by
T
xy
=
24 x24
y x
35
is not a linear transformation. We leave it to the reader to explain why not.
1.9: The Matrix of a Linear Transformation
Denition 74 A transformation T : Vn ! Vm is said to be onto Vm if therange of T is Vm.
Denition 75 (Alternate Denition) A transformation T : Vn ! Vmis said to be onto Vm if each vector y 2 Vm is the image of at least onevectorx 2 Vn.Denition 76 A transformation T : Vn ! Vm is said to be onetooneif each vectory 2 Vm is the image of at most one vectorx 2 Vn.Denition 77 A transformation T : Vn ! Vm that is both onto Vm andonetoone is called a bijection.
Remark 78 In reference to the dart analogy that we have been using in
class: Supposing that Vn is our box of darts and Vm is our dartboard, atransformation T : Vn ! Vm is onto Vm if every point on the dartboardgets hit by at least one dart; whereas T is onetoone if every point on thedartboard either does not get hit or gets hit by at most one dart. If T is abijection, then every point on the dartboard gets hit by exactly one dart.
2.2: The Inverse of a Matrix
Denition 79 Suppose that A =
a bc d
is a 2 2 matrix. The quantity
ad
bc is called the determinant ofA. It is denoted either by det(A) or byjAj.Denition 80 The n n identity matrix, denoted by In, is the matrixthat has entries of 1 at all positions along its main diagonal and entries of 0elsewhere.
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Example 81 The3 3 identity matrix is
I3 =
24 1 0 00 1 0
0 0 1
35 .
Denition 82 An n n matrix, A, is said to be invertible if there existsan n n matrix, B, such that BA = AB = In.
Proposition 83 If A is an invertible n n matrix and B and C are n nmatrices such that BA = AB = In and CA = AC = In, then B = C. Inother words, if A has an inverse, then this inverse is unique.
Denition 84 Suppose that A is an invertible n n matrix. The inverseof A is dened to be the unique matrix, A1, such that A1A = AA1 = In.
Example 85 The inverse of the matrix
A =
3 01 3
is
A1 =
1
30
1
9
1
3
.
(The reader should perform the computations to check that A1A = AA1 =I2.)
Denition 86 An elementary matrix is a matrix is a matrix that canbe obtained by performing a single elementary row operation on an identitymatrix.
Example 87 The matrix
E =24
1 0 00 1 09 0 1
35
is an elementary matrix because it is obtained from I3 by the elementary rowoperation9 R1 + R3 ! R3.
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2.3: Characterizations of Invertible Matrices
Denition 88 A transformation T : Vn ! Vn is said to be invertible ifthere exists a transformation S : Vn ! Vn such that S(T(x)) = x for allx 2 Vn and T (S(x)) = x for allx 2 Vn.
Proposition 89 If T : Vn ! Vn is invertible, then there is a unique trans-formation S : Vn ! Vn such that S(T(x)) = x for all x 2 Vn andT (S(x)) = x for all x 2 Vn. This unique transformation is called theinverse of T and is denoted by T1.
Example 90 If T : V2 ! V2 is the linear transformation that rotates allvectors in V2 counterclockwise about the origin through an angle of 90
, thenT is invertible and T1 : V2 ! V2 is the linear transformation that rotatesall vectors in V2 clockwise about the origin through an angle of 90.
Theorem 91 Suppose that A is an n n matrix and let T : Vn ! Vn bethe linear transformation dened by T(x) = Ax for all x 2 Vn. Then the
following statements are all equivalent to each other (meaning that all of thestatements are true or all of them are false).
1. A is invertible.
2. T is invertible.3. T is onto Vn.
4. T is onetoone.
5. A In.6. The columns of A span Vn.
7. The homogeneous equation Ax = 0n has only the trivial solution.
8. For everyb
2Vn, the equation Ax = b has a unique solution.
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