Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 3.C.7

Exercise 3.C.7

Answers

Proof. Use the same choice of basis for $M (S)$ , $M (T)$ and $M (S + T)$ and same notation as in 3.32.

Let $A = M (S)$ , $B = M (T)$ and $C = M (S + T)$ . Then

\begin{aligned} \sum_{j = 1}^{m} C_{j, k} w_{j} & = (S + T) v_{k} \\ = S v_{k} + T v_{k} \\ = \sum_{j = 1}^{m} A_{j, k} w_{j} + \sum_{j = 1}^{m} B_{j, k} w_{j} \\ = \sum_{j = 1}^{m} (A_{j, k} + B_{j, k}) w_{j} . \end{aligned}

Since $w_{1}, \dots, w_{m}$ is a basis, by 2.29 their coefficients are unique for each vector they determine, therefore it follows that $C_{j, k} = A_{j, k} + B_{j, k}$ , as desired. □

celiopassos

2017-10-06 00:00

Exercise 3.C.7

Answers

Comments

Add answer