An Exercise

• Jan 14th 2011, 10:35 PM
sshh
An Exercise
• Jan 14th 2011, 10:36 PM
dwsmith
Quote:

Originally Posted by sshh

Do you just want the transpose of the matrix?
• Jan 14th 2011, 11:05 PM
FernandoRevilla
Using well known properties of transpose matrices, it is easy to prove $p(A^T)=(p(A))^T$

Fernando Revilla
• Jan 15th 2011, 08:44 AM
HallsofIvy
This is simply "arithmetic".

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

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

Or, using fernandorevilla's suggestion, calculate
$\rho(A)= \begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^3- 5\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}^2+ 11\begin{bmatrix}1 & 3 \\ -1 & 0\end{bmatrix}+ \begin{bmatrix} 4 & 0 \\ 0 & 4\end{bmatrix}$
and then take the transpose.
• Jan 15th 2011, 09:47 AM
FernandoRevilla
Quote:

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

We needn't calculate anything. By hypothesis:

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

so,

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

That is all. :)

Fernando Revilla