Eigenvalues of $A^TA$ 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 $A$.
 A: Counterexample:
$$\begin{align*}
\operatorname{eig} \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} &= \{0,1\} \\
\operatorname{eig} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} &= \{0,2\} \\
\end{align*}$$
A: A real normal matrix is the matrix that satisfies $AA^T = A^T A$. That's what wiki says on normal matrices

Among complex matrices, all unitary, Hermitian, and skew-Hermitian matrices are normal.

Skew-Hermitan matrices are promising for counterexample, since their eigenvalues are purely imaginary. Real skew-Hermitan matrix is just a skew-symmetrical one.
Let $A$ be the skew-symmetrical matrix. Consider eigenvector $x$ of $A$. He have
$$
A^TA x = -A^2 x = -\lambda^2 x
$$
A: 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.
