so $A$ is a $nxn$ matrix with minimal polynomial $f(x)=x^5+x^3+2x+1$ but I need to find the minimal polynomial of $I+A^{-1}$.

What I've worked out so far is that $f(A)=0$ since it's the minimal polynomial, therefore $0=A^5+A^3+2A+I$, so then this leads to $I=A(-A^4-A^2-2I)$ and that $A^{-1}=-A^4-A^2-2I$, showing that $A$ is invertible.

So now letting $B=I+A^{-1} \Rightarrow A=(B-I)^{-1}$ since $f(A)=0 \Rightarrow f((B-I)^{-1})=0$

This leads to $(B-I)^{-5}+(B-I)^{-3}+2(B-I)^{-1}+I=0$ when put into the earlier $f(x)$ statement. This would then lead to $I=(B-I)^{-1}[-(B-I)^{-4}-(B-I)^{-2}-2I]$ so that $B=-(B-I)^{-4}-(B-I)^{-2}-2I$.

I am just having trouble trying to figure out what I would do next to figure out the minimal polynomial of $I+A^{-1}$ exactly. Any help would be appreciated.

  • $\begingroup$ Since $f$ is a polynomial with $5$ different complex roots, $A$ is diagonalisable, and invertible as well since $0$ is not a root of $f$. Therefore, $I+A^{-1}$ is diagonalisable, and it is easy to find its minimal polynomial. $\endgroup$ – Gabriel Romon Mar 13 '15 at 5:44
  • $\begingroup$ You made an error in your deduction; see the update of my answer. $\endgroup$ – Marc van Leeuwen Mar 13 '15 at 8:24

You made an error in deducing the equation for $B$: it should read, with an additional $-I$ on the left: $B-I=-(B-I)^{-4}-(B-I)^{-2}-2I$.

From this relation, rewritten $(B-I)^{-4}+(B-I)^{-2}+I+B=0$, you just need to make all powers positive. So multiply by $(B-I)^4$ to get $I+(B-I)^2+(B-I)^4(I+B)=0$ and expand.

A somewhat more direct, but essentially equivalent deduction would be as follows. From the fact that the minimal polynomial $x^5+x^3+2x+1$ has nonzero constant terms, conclude (as you did) that $A$ is invertible, and then that $A^{-1}$ has a "reverse" minimal polynomial: multiplying $A^5+A^3+2A+1=0$ by $A^{-1}$ gives $1+A^{-2}+2A^{-4}+A^{-5}=0$ so $x^5+2x^4+x^2+1$ annihilates $A^{-1}$. And it must be of minimal degree doing so, as any proper divisor would also have nonzero constant term and similarly lead to a lower degree polynomial annihilating $A$ which is absurd. To get from the minimal polynomial of $A^{-1}$ to that of $A^{-1}+1$, where it will be convenient to write the latter in terms of a new indeterminate $y$, note that substituting $A^{-1}+1$ for $y$ can be done by first substituting $x+1$ for $y$ (an obviously invertible operation) and then substituting $A^{-1}$ for $x$. We saw that the kernel of the latter operation is the ideal generated by $x^5+2x^4+x^2+1$, so the kernel of the whole operation is the ideal generated by the polynomial mapping to it under the first substitution, which is $(y-1)^5+2(y-1)^4+(y-1)^2+1$.


You alrady found that $B=I+A^{-1} = -A^4-A^2-I$. One way would be to calculate $B^2$, $B^3$, $B^4$ as polynomials in $A$ (always reducing modulo $f(A)$ so that all expressions are of degree $\le 4$); then find a linear dependence between these expressions


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