2.9. EXERCISES 79

35. You have a linear transformation T and

T

 1

2

−18

 =

 5

2

5

 , T

 −1

−1

15

 =

 3

3

5

T

 0

−1

4

 =

 2

5

−2

Find the matrix of T . That is find A such that Tx = Ax.

36. Suppose V is a subspace of Fn and T : V → Fp is a nonzero linear transformation.Show that there exists a basis for Im (T ) ≡ T (V )

{Tv1, · · · , Tvm}

and that in this situation,{v1, · · · ,vm}

is linearly independent.

37. ↑In the situation of Problem 36 where V is a subspace of Fn, show that there exists{z1, · · · , zr} a basis for ker (T ) . (Recall Theorem 2.6.12. Since ker (T ) is a subspace,it has a basis.) Now for an arbitrary Tv ∈ T (V ) , explain why

Tv = a1Tv1 + · · ·+ amTvm

and why this implies

v − (a1v1 + · · ·+ amvm) ∈ ker (T ) .

Then explain why V = span (v1, · · · ,vm, z1, · · · , zr) .

38. ↑In the situation of the above problem, show {v1, · · · ,vm, z1, · · · , zr} is a basis for Vand therefore, dim (V ) = dim (ker (T )) + dim (T (V )) .

39. ↑Let A be a linear transformation from V to W and let B be a linear transformationfrom W to U where V,W,U are all subspaces of some Fp. Explain why

A (ker (BA)) ⊆ ker (B) , ker (A) ⊆ ker (BA) .

ker(B)

A(ker(BA))

ker(BA)

ker(A)A

40. ↑Let {x1, · · · ,xn} be a basis of ker (A) and let

{Ay1, · · · , Aym}

be a basis of A (ker (BA)). Let z ∈ ker (BA) . Explain why

Az ∈ span {Ay1, · · · , Aym}