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The following questions is apprently an extentsion of cayley hamilton and I was not sure if that meant we where supposed to use the proof of said result but I do not know how to extend the proof using Hofman and Kunze's text.

Suppose that $p_M$ is the characteristic polynomial for a matrix $M$ with entries in a field. Suppose there exists a matrix $N$ such that $MN = NM$.

How do we show that $ p_M (N) = A (M - N) $ for some matrix $A$ such that $A$ commutes with $M$ and $N$?

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You introduce $p_N$ but then use $P_M$ -- are these supposed to be the same? – joriki Nov 17 '11 at 11:15
Thank you it was supposed to be $P_M$. – user7980 Nov 17 '11 at 11:29
You've changed the $N$ to an $M$, but now you have $p_M$ and $P_M$. Again, are these supposed to be the same? – joriki Nov 17 '11 at 11:35
up vote 2 down vote accepted

Write $p_M(x) = \sum_{i=0}^n c_ix^i$. And since $p_M$ is the char. poly. of $M$, we have $p_M(M) = 0$. Observe:

$$ \begin{eqnarray*} p_M(N) &=& p_M(N) - p_M(M) \\ &=& \sum_{i=0}^n c_iN^i - \sum_{i=0}^n c_iM^i\\ &=& \sum_{i=1}^n c_i(N^i- M^i)\\ \end{eqnarray*} $$ Note, I can start the indexing at $i=1$ since for $i=0$, we get $c_0-c_0=0$. Continuing on, notice that each $N^i-M^i$ factors as $Q_i(N-M)$, where $Q_i$ is a polynomial of degree $i-1$ in $N$ and $M$. This factorization only works because $N$ and $M$ commute with one another (this point is crucial). Moreover, both $M$ and $N$ commute with $Q_i$ (why?). $$ \begin{eqnarray*} p_M(N) &=& \sum_{i=1}^n c_iQ_i(N-M) \\ &=& \left[\sum_{i=1}^n c_iQ_i\right](N-M) \end{eqnarray*} $$ Thus, the matrix $A=\sum_{i=1}^n c_iQ_i$.

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