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

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

• If $A$ is the zero matrix, then $f(A)$ will just be the constant term of $f$, which must be a scalar matrix. So in this case, it is impossible. – Joppy Sep 26 '18 at 5:08
• @Joppy Sir In case if I assume non zero matrix then is it possible? – MathLover Sep 26 '18 at 5:13
• This will only be possible if $\begin{bmatrix} 0 & -1\\1 & 0 \end{bmatrix}$ is a linear combination of the powers of $A$. So you need to look at the powers of $A$ to decide. (It is not always possible) – Morgan Rodgers Sep 26 '18 at 5:44
• It is easy to see that $f(A)$ and $A$ commute. You only need a matrix that does not commute with $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and you have a counterexample. – Reinhard Meier Sep 26 '18 at 6:54

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: $$f(x) = \frac xc - \frac ac$$