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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

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  • $\begingroup$ 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. $\endgroup$ – Joppy Sep 26 '18 at 5:08
  • $\begingroup$ @Joppy Sir In case if I assume non zero matrix then is it possible? $\endgroup$ – MathLover Sep 26 '18 at 5:13
  • $\begingroup$ 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) $\endgroup$ – Morgan Rodgers Sep 26 '18 at 5:44
  • $\begingroup$ 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. $\endgroup$ – Reinhard Meier Sep 26 '18 at 6:54
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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: $$ f(x) = \frac xc - \frac ac $$

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