Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 10.A.4

Exercise 10.A.4

Answers

Proof. Let $A = M (T, (v_{1}, \dots, v_{n}))$ . Then

I u_{k} = T v_{k} = A_{1, k} v_{1} + \dots + A_{n, k} v_{n} .

The equation above shows that $M (I, (u_{1}, \dots, u_{n}), (v_{1}, \dots, v_{n})) = A$ , because each $I u_{k}$ can be uniquely written as linear combination of the $v$ ’s and this combination forces the desired equality. □

celiopassos

2017-10-06 00:00

Exercise 10.A.4

Answers

Comments

Add answer