Exercise 1.6.11

The eigenvalues of $A$ equal the eigenvalues of $A^T$. This is because the determinant of a matrix $A$ equals to the determinant of its transpose $A^T$, so $\mathit{det}(A-\mathit{\lambda I})=\mathit{det}{(A-\mathit{\lambda I})}^T=\mathit{det}(A^T-\mathit{\lambda I})$.

Example, for $A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

The eigenvalues of $A$ are: ${\lambda }_1=2$, ${\lambda }_2=-1$. The eigenvectors are $v_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$, and $v_2=\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 1\hfill \end{array}\right]$.

The eigenvalues of $A^T$ are: ${\lambda }_1=2$, ${\lambda }_2=-1$. The eigenvectors are $v_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$, and $v_2=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$.