"proof" A is a Hermitian Matrix 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.
 A: 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.
A: 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.
