Exercise 5.A.24

Answers

Proof. (a) It is easy to see that $M (T, e_{1}, \dots, e_{n}) = A$ . Thus

T e_{k} = \sum_{j = 1}^{n} A_{j, k} e_{j} .

and then

\begin{aligned} T (e_{1} + \dots + e_{n}) & = \sum_{j = 1}^{n} A_{j, 1} e_{j} + \dots + \sum {j = 1}^{n} A_{j, n} e_{j} \\ = (\sum_{j = 1}^{n} A_{1, j}) e_{1} + \dots + (\sum_{j = 1}^{n} A_{n, j}) e_{n} \\ = e_{1} + \dots + e_{n} . \end{aligned}

Hence $e_{1} + \dots + e_{n}$ is an eigenvector of $T$ corresponding to the eigenvalue $1$ . □

celiopassos

2017-10-06 00:00