rank one pertubations of the identity, and a matrix equation [closed]

Let $u$, $v$ be two orthogonal vectors in $\mathbb C^n$, such that $u\cdot v = 0$.

Notice if $A = I + uv$, then $A^{-1} = I – uv$, thus $A + A^{-1} = 2I$.

Consider the equation:

$$A + A^{-1} = 2I.$$

Are there any other solutions to this equation?

I am inclined to say, no.

I also asked this on my blog:

http://ryanlewis.net/p/challenge-problem

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A matrix satisfying that equation also satisfies the polynomial $(X-1)^2$. If the matrix is not a scalar matrix, then this is then its minimal polynomial. It is easy to see what the possible Jordan forms of $A$ are from this information, so up to conjugacy we know all solutions. Your question can be answered using it. – Mariano Suárez-Alvarez Oct 31 2010 at 16:32
(In particular, you see from this we see that $$\left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array} \right)$$ is a counterexample) – Mariano Suárez-Alvarez Oct 31 2010 at 16:35
Mariano has explained this very well. This question is more suitable for math.stackexchange.com than here. – Deane Yang Oct 31 2010 at 17:09
Try $A=I+R$ where $R^2=0$. As Mariano's example shows, $R$ need not be a rank-one matrix. – Yemon Choi Oct 31 2010 at 22:57