For an arbitrary complex matrix A show that $$A*A^\dagger$$ is Hermitian.
Where the dagger "$\dagger$" stands for the "complex conjugate and transpose" operators.
From what I understand this must mean that $$A*A^\dagger = [A*A^\dagger]^\dagger$$ But I am stuck. I don't really understand the properties of the complex conjugate function with Matrices.
On
This is perhaps different than the way it's presented in physics class, but it's better and generalizes easily to infinite dimensional spaces and non-Hilbert spaces. Use the mathematicians definition of adjoint: $A^t$ is adjoint to $A$ iff $(A u,v)=(u,A^t v)$ for all possible $u,v$.
Then bringing the operators over one at a time and using associativity of matrix multiplication yields:
$$(AA^t u,v)=(A^t u,A^t v)=(u,AA^t v)$$
So by the above definition $AA^t$ is adjoint to itself.
The key properties you need are the fact that hermitian-conjugating twice returns you to where you started, $$(A^\ast)^\ast=A$$ (which holds because it holds for both transposing and complex-conjugating), and the formula for the conjugate of a product, $$(AB)^\ast=B^\ast A^\ast,$$ which is inherited from the identical transpose formula and is unchanged by complex conjugation.
Of course, you can also prove both formulas from the (real) definition of the conjugate of $A$, namely that for all $u$ and $v$ you get $\langle A u,v\rangle=\langle u, A^\ast v\rangle$, as Nick points out.