Eigenvalues of $A^TA$

Much of this is made much easier to see if you know about the spectral theorem. My answers below rely heavily on this.

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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 $|\lambda_k|^2$.

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 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 $$ A^TA x = -A^2 x = -\lambda^2 x $$