# Proving an orthogonal matrix minus the identity matrix is invertible

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!

• 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. – user856 Oct 28 '19 at 18:49

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