Exercise 1.6.25

Since the eigenvectors in $A^{T}y=\mathit{\lambda y}$ are the columns of that matrix ${(X^{-1})}^T$, so we have $Y=\left[\begin{array}{cccc}\hfill y_1\hfill & \hfill y_2\hfill & \hfill \dots \hfill & \hfill y_n\hfill \end{array}\right]={(X^{-1})}^T$, thus $X^{-1}=\left[\begin{array}{c}\hfill y_1^T\hfill \\ \hfill y_2^T\hfill \\ \hfill \dots \hfill \\ \hfill y_n^T\hfill \end{array}\right]$.

Note, $x_i$ and $y_i$ are all column vectors here.

Thus $A=\mathit{X\Lambda }{X}^{-1}=\mathit{X\Lambda }\left[\begin{array}{c}\hfill y_1^T\hfill \\ \hfill y_2^T\hfill \\ \hfill \dots \hfill \\ \hfill y_n^T\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill {\lambda }_1{x}_1\hfill & \hfill \dots {\lambda }_n{x}_n\hfill \end{array}\right]\left[\begin{array}{c}\hfill y_1^T\hfill \\ \hfill y_2^T\hfill \\ \hfill \dots \hfill \\ \hfill y_n^T\hfill \end{array}\right]={\lambda }_1{x}_1{y}_1^T+\cdots +{\lambda }_n{x}_n{y}_n^T$