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Let $ A, X \in M_{nxn}(K) $. Let $ p_A(t) $ be a characteristic polynomial of matrix A. Let's assume that $ XA = AX $. Show that there is such a matrix $M$ that $ p_A(X) = M(A-X), MA=AM$ and $ MX=XM $.

I believe I need to use Jordan form in order to proceed. I could consider two scenarios: when X and A are invertible and thus similar and when these two are not invertible. Is this the right approach?

Any hints?

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Write $p_A(t)=\sum_{k=0}^n a_k t^k$, the characteristic polynomial of $A$. By Cayley-Hamilton theorem, $p_A(A)=0$.

Then we have $$ \begin{align} p_A(X)&=p_A(X)-p_A(A)\\ &=\sum_{k=0}^n a_k (X^k-A^k). \end{align} $$ Since $XA=AX$, we have for $k\geq 1$, $$ X^k-A^k=(X-A)\sum_{i=0}^{k-1} X^iA^{k-1-i}. $$

Hence, by noting that $A^0=X^0=I$ for convenience, $$\begin{align} p_A(X)&=(X-A)\sum_{k=1}^n a_k\sum_{i=0}^{k-1}X^iA^{k-1-i} \end{align} $$

Then take $$ M=-\sum_{k=1}^n a_k\sum_{i=0}^{k-1}X^iA^{k-1-i}. $$

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