1. ## An Exercise

Consider
If compute

2. Originally Posted by sshh
Consider
If compute
Do you just want the transpose of the matrix?

3. Using well known properties of transpose matrices, it is easy to prove $\displaystyle p(A^T)=(p(A))^T$

Fernando Revilla

4. This is simply "arithmetic".

With $\displaystyle A= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}$,
$\displaystyle A^*= \begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}$ so that

$\displaystyle \rho(A^*)= \begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & -1 \\ 3 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}$
Have you done that? With a 2 by 2 matrix, you should find the calculation fairly easy.

Or, using fernandorevilla's suggestion, calculate
$\displaystyle \rho(A)= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}$
and then take the transpose.

5. Originally Posted by HallsofIvy
Or, using fernandorevilla's suggestion, calculate
$\displaystyle \rho(A)= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}$
and then take the transpose.
We needn't calculate anything. By hypothesis:

$\displaystyle p(A)=\begin{bmatrix}{\;\;\;1}&{3}\\{-1}&{0}\end{bmatrix}$

so,

$\displaystyle p(A^T)=(p(A))^T=\begin{bmatrix}{1}&{-1}\\{3}&{\;\;\;0}\end{bmatrix}$

That is all.

Fernando Revilla