Proving that a normal matrix $\mathbf{A}$ and $\mathbf{A'A}$ has same eigenvectors

Question

Is my proof correct ?.

$\mathbf{A}$ be the normal matrix, and $x$ is the eigenvector corresponding to the eigenvalue $\lambda$, then $$ \begin{align} \mathbf{Ax} &=\lambda x\\ \mathbf{A'Ax} =& \mathbf{A'}\lambda x = \lambda \mathbf{A'} x\\ & \implies \lambda^2 x \end{align} $$ Hence $\mathbf{x}$ is an eigenvector with eigenvalue $\lambda^2$

Answer 1

You only proved one inclusion.

In other words, you proved that if $A$ is normal, then every eigenvector of $A$ is also an eigenvector of $A^{*}\!A$.

However, the reverse inclusion need not hold.

For example, if $A$ is given by $$ A= \pmatrix{ 0&1\\ 1&0\\ } $$ then $A$ is real-symmetric, hence also normal.

But $A^{*}\!A=I$, so every nonzero vector is an eigenvector of $A^{*}\!A$, but not every nonzero vector is an eigenvector of $A$.