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Selected SolutionsMath 420

Homework 52/10/12

3.1.3 Describe the range and the null space for the differentiation transformation ofExample 2. Do the same for the integration transformation of Example 5.

For the differentiation transformation of Example 2, the null space is the set of constantfunctions f(x) = c0, since a function whose derivative is identically zero must be con-stant. The range is all ofV, since every polynomial is the derivative of its indefiniteintegral.

For the integration transformation of Example 5, the null space consists of only thezero function f(x) = 0, since the integral of a nonzero function from 0 to x is nonzeroas a function ofx. The range consists of all continuously differentiable functions with

f(0) = 0, since in this case f(x) =x

0 f(t)dt, and f(0) = 0 is necessary since f(0) =00 g(t)dt = 0 for some g.

3.1.4 Is there a linear transformationTfrom R3 to R2 such thatT(1,1, 1) = (1, 0)andT(1,1,1) = (0, 1)?

Yes. Clearly(1,1, 1)and(1,1,1)are linearly independent, so there exists a basis con-taining these vectors by Theorem 5 in Chapter 2. Thus, by Theorem 1, there exists a lineartransformation withT(1,1, 1) = 1andT(1,1,1) = 2for any1,2in any vector spaceover R. In particular, we can chooseT(1,1, 1) = (1, 0)andT(1,1,1) = (0, 1).

3.1.5 If

1 = (1,1), 1=(1, 0)

2 = (2,1), 2=(0, 1)

3 = (3, 2), 1=(1, 1)

is there a linear transformation Tfrom R2 into R2 such thatTi = ifori = 1, 2 and 3?

No. If such a Texists, we have T(3

) = T(1

2) = T(

1) T(

2) = (1, 0)

(0, 1) = (1,1), which is a contradiction since we are givenT(3) = (1, 1).

3.2.2 LetTbe the (unique) linear operator on C3 for which

T1= (1,0, i), T2 = (0,1,1), T3= (i, 1 , 0).

IsTinvertible?

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No. T(i, 1,1) = i(1,0, i) + (0,1,1) (i, 1 , 0) = 0, soTis not one-to-one, and hence notinvertible.

3.2.5 Let C22 be the complex vector space of 2 2 matrices with complex entries. Let

B=

1 14 4

and letTbe the linear operator on C22 defined byT(A) = BA. What is the rank ofT?Can you describeT2?

Let

A1 =

1 00 0

, A2=

0 10 0

, A3=

0 01 0

, andA4 =

0 00 1

.

Clearly, {A1,A2,A3,A4} is a basis ofC22. We have

T(A1) =

1 04 0

, T(A2) =

0 10 4

,

T(A3) =

1 0

4 0

= T(A1), and T(A4) =

0 10 4

= T(A2).

Thus, the range ofTis spanned byT(A1)andT(A2), which are clearly linearly indepen-dent, and hencerank(T) =2.

We have T2(A) = T(T(A)) = T(BA) = BBA = B2A. Straightforward computationshows thatB2 =5B, so T2(A) =5T(A).

3.2.6 LetTbe a linear transformation from R3 into R2, and letUbe a linear transfor-mation from R2 into R3. Prove that the transformationUTis not invertible. Generalizethe theorem.

By Theorem 2,rank(T) +nullity(T) =dim(R3) =3. Since the range ofTis a subspace ofR2,rank(T) 2, sonullity(T) > 0 and there exists a nonzero x R3 withTx = 0. Wehave(UT)x= U(Tx) =U0= 0, soUTis not one-to-one and not invertible.

In general, ifT : V WandU : W Vare linear transformations and dim(W)