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!
Best Answer
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.