TITLE: Eigenvalues of $A^TA$
QUESTION [12 upvotes]: Suppose $A$ is a $n\times n$ matrix in $M(\mathbb{R})$. I'd like to know if there is an exact formula for the eigenvalues of $A^TA$. Clearly, it's false that $\lambda^2$ is an eigenvalue of $A^TA$ if $\lambda$ is an eigenvalue of $A$.

REPLY [10 votes]: Much of this is made much easier to see if you know about the spectral theorem.  My answers below rely heavily on this.

For your first statement, any non-normal matrix provides a counterexample.  In fact, we have the following theorem:

Let $A$ be a square matrix with eigenvalues $\lambda_k$.  Let $\sigma_1,\dots,\sigma_n$ denote the eigenvalues of $A^TA$ (which are all positive).
Then
$$
\sum_{k=1}^n |\lambda_k|^2 \leq  \sum_{k=1}^n \sigma_k
$$
and $A$ is normal if and only if
$
\sigma_k  = |\lambda_k|^2
$ for each $k$.

The proof of your second statement (by the spectral theorem) is as follows:
Because $A$ is normal, there exists a unitary matrix $U$ and diagonal matrix $D$ (each with complex entries) such that
$A = UDU^*$
where $M^* = \overline{M^T}$ denotes the conjugate-transpose, AKA the adjoint of a complex matrix.  Note that
$$
D = \pmatrix{\lambda_1\\&\ddots \\&& \lambda_n}
$$
where $\lambda_k$ are the eigenvalues of $A$.
We then have
$$
A^TA = A^*A = (UDU^*)^*UDU^* = 
UD^*DU^* = U \pmatrix{|\lambda_1|^2\\ & \ddots \\ && |\lambda_n|^2}U^*
$$
Thus, the eigenvalues of $A^TA$ are $|\lambda_k|^2$.

As noted in the comments on the question, there is no formula for the eigenvalues of $A^TA$ that only uses the eigenvalues of $A$. This post addresses the relationship between these quantities in general.