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Let $M$ be an invertible $n \times n$ matrix whose minimal polynomial $q_M$ and characteristic polynomial are equal. Suppose as well that the $M$ has entries in an algebraically closed field.

How do we show that the minimal polynomial of $M^{-1}$ is given by $q_M(0)^{-1} x^n q_M(\frac{1}{x}) $? ($x$ is the indeterminate in the polynomial ring over the base field $k$?

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You're aware of the relationship between the roots of a polynomial $p(x)$ and the "reciprocal polynomial" $x^n p(1/x)$? Note that the extra factor is needed so that the minimal polynomial is monic (as a hint, why is $p(0)$ the same as the constant term of $p(x)$?). – J. M. Dec 5 '11 at 8:10
It would seem that the minimal polynomial being equal to the characteristic polynomial is a red herring: it isn't needed. Also your title is strange, as it provokes the reply "well, then compute the characteristic polynomial!". – Marc van Leeuwen Dec 5 '11 at 12:44
up vote 2 down vote accepted

Let $f(x) = q_M(0)^{-1} x^n q_M(1/x)$. Note that $f$ is a polynomial of degree $n$ and $f(M^{-1}) = 0$. Similarly, if $g(M^{-1}) = 0$ where $g$ is a polynomial of degree $k < n$ with nonzero constant term, then $h(x) = x^k g(1/x)$ would be a polynomial of degree $k$ such that $h(M) = 0$, contradicting minimality of $q_M$.

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