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Would appreciate help with the following problem:

Given an invertible $n \times n$ real matrix $A$ with eigenvalue $t$ of geometric multiplicity at least $2$, determine whether the following statement is true: "There is a real polynomial $Q(x)$ of degree smaller or equal to $n-2$ such that the inverse of $A$ equals $Q(A)$".

It's easy to prove a similar statement, that the inverse of A can be represented by an $n-1$ degree polynomial (if I am correct, this is true for any invertible matrix), but that's as far as I got. I was thinking maybe the Jordan form of $A$ can help me out, but I haven't gotten very far with that. Would appreciate hints!

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If $\mu(x)$ is the characteristic polynomial, then $\mu(A)=0$; since $A$ is invertible, $\mu(0)\neq 0$, so you can write $1 = -\frac{1}{\mu(0)}xm(x)$, where $m(x) = (\mu(x)-\mu(0))/x$. Plugging $A$ tells you that you can always express $A^{-1}$ as a polynomial of degree at most $n-1$ in $A$. But you can play the same game with the minimal polynomial, and the fact that the geometric multiplicity of $t$ is at least $2$ tells you something about the largest degree of $(x-\mathbf{t})$ that can divide the minimal polynomial. Play with that and see if you can get the desired conclusion. – Arturo Magidin Sep 11 '11 at 23:16
up vote 1 down vote accepted

To expand on my comment: A typical way of showing that if $A$ is invertible then there is a polynomial $q(x)$ such that $A^{-1}=q(A)$ is to use Cayley-Hamilton: writing $$\chi(x) = x^n + a_{n-1}x^{n-1}+\cdots+a_0$$ for the characteristic polynomial of $A$, we know that $a_0$ is the product of the eigenvalues, and so $A$ is invertible if and only if $a_0\neq 0$. The Cayley-Hamilton Theorem tells us that $\chi(A)=0$, so $$0 = A^n + a_{n-1}A^{n-1}+\cdots + a_1A + a_0I,$$ from which we get that $$-a_0I = A^n+a_{n-1}A^{n-1}+\cdots + a_1A,$$ hence $$I =-\frac{1}{a_0}A\left( A^{n-1} + a_{n-1}A^{n-2}+\cdots + a_1I\right).$$ Thus, putting $q(x) = -\frac{1}{a_0}\left(x^{n-1}+a_{n-1}x^{n-2}+\cdots + a_1\right)$ gives the desired polynomial.

Of course, you can play the exact same game with the minimal polynomial instead of the characteristic polynomial. The matrix $A$ is invertible if and only if $0$ is not an eigenvalue, if and only if $x$ does not divide the minimal polynomial. The geometric multiplicity of $\mathbf{t}$ tells you something about the degree of the minimal polynomial: since the highest degree of $x-\mathbf{t}$ that divides the minimal polynomial is the size of the largest Jordan block associated to $\mathbf{t}$, and since the geometric multiplicity equals the number of Jordan blocks, if the algebraic multiplicity of $\mathbf{t}$ is, say, $m$, having geometric multiplicity $2$ tells you that there are at least two Jordan blocks associated to $\mathbf{t}$, and so the largest block will have size at most $m-1$; which tells you that if the largest power of $x-\mathbf{t}$ that divides $\chi(x)$ is $m$, then the largest power of $x-\mathbf{t}$ that divides the minimal polynomial is at most $m-1$.

Can you leverage that into showing that a $Q(x)$ as you want must exist?

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Very helpful, thank you! I was working the details from when you posted your comment. I think I have it worked out. Right now, I need to catch up on my sleep, but I'll post my proof (using your hints) here tomorrow, to see if I got it right. – josh Sep 11 '11 at 23:35
Well, though if I have it right, there is very little work to be done at this point (I think we get that the minimal polynomial is of degree at most n-1, which, using the same trick you've shown, yields a polynomial of degree <= n-2 as required). In any case, good night. – josh Sep 11 '11 at 23:44
@ArturoMagidin - how can we ensure that $q(x)$ is a real polynomial? – trfv Sep 15 '15 at 11:10

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