$A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^TA$ is diagonalisable? I have been set some work to do over the holidays, and one of the questions gives a hint that is as follows:
$A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^TA\text{ is diagonalisable}$.
I know that


*

*$A\in\mathrm{M}_{n\times n}(\mathbb{R})\implies A^TA\text{ is diagonalisable}$

*$A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^*A\text{ is diagonalisable}$
where the former can be thought of as a particular case of the latter. Both those statements are true because $A^*A$ is self adjoint, and we can then apply the Spectral Theorem for normal operators.
But is the statement at the top of my question true, or has the lecturer simply mistyped one of the two facts that I've written? If it is true, I can't see how to prove it, so any hints would be appreciated.

I think that's a typo. Your lecturer should write $A^\ast A$ instead of $A^TA$.
In general, every complex square matrix $C$ is $\mathbb C$-similar to a complex symmetric matrix $S$. Since a symmetric matrix represents a symmetric bilinear form, and every symmetric bilinear form is diagonalisable (via $T$-congruence), $S$ can always be written as $A^TA$ for some complex matrix $A$. So, if what your lecturer wrote is correct, that means every complex square matrix $C$ is diagonalisable. Yet this is obviously untrue.
For a concrete example, consider
$$
A=\pmatrix{1&\tfrac{i+1}2\\ 1&\tfrac{i-1}2},\ A^TA=\pmatrix{2&i\\ i&0}=
\pmatrix{-i&-i\\ 1&0}\pmatrix{1&1\\ 0&1}\pmatrix{0&1\\ i&-1}.
$$
Since the Jordan form of $A^TA$ is $\pmatrix{1&1\\ 0&1}$, $A^TA$ is nondiagonalisable.
