If $A \in \mathbb{R}^{m\times n}$, $m<n$, and $AA^T = I_m$, what are the eigenvalues of $3I_n - A^TA$?

Question

If $A \in \mathbb{R}^{m\times n}$, $m<n$, and $AA^T = I_m$, what are the eigenvalues of $3I_n - A^TA$?

How can I start? I know the eigenvalue of $I$ is $1$, but how do I find the eigenvalues of $A^TA$?

Answer 1

Saying that $\lambda$ is an eigenvalue of $3I_n-A^TA$ is the same as saying that $-\lambda$ is an eigenvalue of $A^TA-3I_n$ or, as well, that $$ A^TA-3I_n+\lambda I_n=A^TA-(3-\lambda)I_n $$ is not invertible. In other words, the eigenvalues you are looking for are of the form $3-\mu$, where $\mu$ is an eigenvalue of $A^TA$.

Now suppose $\mu\ne0$ is an eigenvalue of $A^TA$ and let $v$ be an eigenvector. Then $A^TAv=\mu v$ (in particular $Av\ne0$), so also $$ (AA^T)(Av)=\mu(Av) $$ so $Av$ is an eigenvalue of $AA^T$ relative to $\mu$. Since $AA^T=I_m$, we conclude $\mu=1$.

Since $A^TA$ is $n\times n$ and $n>m$, we also know that $0$ is an eigenvalue.