For any given matrix Is there exist some polynomial which can convert that into required one?

Question

I wanted to check following possibility .
I have $$ A=\left [\begin{matrix} a & b \\ c & d\\ \end{matrix} \right] $$ Now I wanted to find Polynomial such that
$$ f(A)=\left [\begin{matrix} 0 & -1 \\ 1 & 0\\ \end{matrix} \right] $$.
Is this always possible?
I had done some calculatution but I did not get.
Any Help will be appreciated

Answer 1

From the fact that $f(A)$ and $A$ commute, we conclude that $A$ must commute with $\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}.$ This in turn results in the requirements $a=d$ and $b=-c.$ If $b=c=0,$ then $f(A)$ would only be a multiple of the identity matrix. Therefore, we can also conclude $b=-c\neq 0.$ Now it is easy to find our polynomial: $$ f(x) = \frac xc - \frac ac $$