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I have this problem:

$A$ is an $n \times n$-matrix, its characteristic polynomial is $P(X)=(X-1)^n$. Prove that $A$ is similar to its inverse.

How do you solve it? I really don't know.

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Do you know Jordan canonical form? – Ted Shifrin Jun 28 '13 at 14:26
yes, i know it quite well. – mrprottolo Jun 28 '13 at 14:31
So what are the possible Jordan canonical forms of $A$? What do you deduce about $A^{-1}$? – Ted Shifrin Jun 28 '13 at 14:34
I know that the Jordan canonical form is made up by some blocks with eigenvalue 1. I think the inverse is an upper triangular matrix with (1 -1 1 -1 ...) sequences in the rows. – mrprottolo Jun 28 '13 at 14:45
Yes, but those are similar, by either explicit or theoretical considerations. – Ted Shifrin Jun 28 '13 at 14:47
up vote 2 down vote accepted

Hints. Call the underlying field $\mathbb{F}$. We will use the following fact:

If the characteristic polynomial of a matrix $B\in M_n(\mathbb{F})$ can be factored into linear factors over $\mathbb{F}$ (i.e. if $B$ has a complete set of eigenvalues in $\mathbb{F}$), $B$ is similar to its Jordan form over $\mathbb{F}$.

  1. Show that $A$ is is similar to its Jordan form over $\mathbb{F}$. So, it suffices to prove the problem statement for a single Jordan block.
  2. From now on, suppose $A$ is an $n\times n$ Jordan block. Write $A$ in the form of $I+J$, where $J$ is an $n\times n$ nilpotent Jordan block (i.e. $J$ has ones on its superdiagonal and zeroes elsewhere). Express $A^{-1}$ as a polynomial in $J$. (Hint: consider the formal power series of $\frac1{1+x}$ and note that $J^n=0$.)
  3. Show that $A^{-1}$ is similar to its Jordan form over $\mathbb{F}$.
  4. By considering $\mathrm{rank}(A^{-1}-I)$, show that the Jordan form of $A^{-1}$ is precisely equal to $A$. Hence $A^{-1}$ is similar to $A$ over $\mathbb{F}$.
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