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I am working on the following Linear algebra problem:

Let $A$ be a real skew-symmetric matrix. Recall that the eigenvalues of $A$ are pure imaginaries, and so $A - I$ is invertible. Let $T = (A+I)(A-I)^{-1}$. Prove that $T - I$ is invertible.

The approach I thought I'd take to this problem is, if $T - I$ is invertible, then since $\det(T-I)$ would be given by the product of the eigenvalues of $T - I$, it follows that $0$ should not be an eigenvalue of $T - I$ $\Rightarrow 1$ should not be an eigenvalue of $T$. Thus, we need to show that $1$ cannot be an eigenvalue of $T$.

Since the eigenvalues of $A$ are pure imaginaries, I noted that the eigenvalues of $A-I$ are of the form $\lambda - 1$, where $\lambda = ib (b \in \mathbb{R} )$ is a pure imaginary number. Similarly, the eigenvalues of $A + I$ are of the form $\lambda + 1$. But, I'm not sure how to use these two facts to say something about the eigenvalues of $T$ -- in general, I couldn't find any relationships between the eigenvalues of $AB$ and the eigenvalues of $A$ and the eigenvalues of $B$ for two matrices $A$ and $B$. How can I show that $T$ cannot have an eigenvalue of $1$ ?

Thanks!

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  • $\begingroup$ Show that any eigenvector of $A$ is also an eigenvector of $A+I$, $A-I$, and $(A+I)(A-I)^{-1}$, and find the associated eigenvalues for each. Then the result in Robert Israel's answer readily follows. $\endgroup$ – user856 Oct 28 '19 at 18:49
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Note that $$ T-I=(A+I)(A-I)^{-1}-(A-I)(A-I)^{-1}=2I(A-I)^{-1}=2(A-I)^{-1} $$ Which is invertible by assumption.

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The Spectral Mapping Theorem says the eigenvalues of $(A+I)(A-I)^{-1}$ are the images of the eigenvalues of $A$ under the map $\lambda \mapsto (\lambda + 1)/(\lambda - 1)$.

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