Eigenvalues and eigenvectors of $A$, $A^\dagger$ and $AA^\dagger$ Consider a general complex matrix $A$ satisfying the eigenvalue equation $$AX=\lambda X$$ where $\lambda$ is an eigenvalue corresponding to the nonzero eigenvector $X$. Let us also assume that the eigenvalues of $A$ are all distinct. I have three closely related questions.
How can we show that the eigenvalues of $A$ are complex conjugates of the eigenvalues of $A^\dagger$ (the complex conjugate transpose of $A$)? For this claim, see here.
How are the eigenvectors of $A$ and $A^\dagger$ related?
Do the above conclusions change if some eigenvalues are repeated?
What can we say about the eigenvalues of $h=AA^\dagger$? Since $h$ is hermitian its eigenvalues must be real. But are they also nonnegative?
 A: The eigenvalues of a matrix $\ A\ $ are the roots of its minimal polynomial.  A polynomial is satisfied by $\ A\ $ if and only if its complex conjugate is satisfied by $\ A^\dagger\ $.  Therefore the minimal polynomials of $\ A\ $ and $\ A^\dagger\ $. are complex conjugates of each other, as are their roots.
If $\ \lambda_1\ $ and $\ \lambda_2\ne \lambda_1\ $ are distinct eigenvalues of $\ A\ $, $\ x_1\ $ an eigenvector of $\ A\ $ corresponding to $\ \lambda_1\ $, and $\ y_2\ $ an eigenvector of $\ A^\dagger\ $ corresponding to $\ \lambda_2^*\ $, then
\begin{align}
y_2^\dagger Ax_1&=\lambda_1 y_2^\dagger x_1\\
&=\left(x_1^\dagger A^\dagger y_2\right)^*\\
&=\left(\lambda_2^*x_1^\dagger y_2\right)^*\\
&=\lambda_2y_2^\dagger x_1\ .
\end{align}
Therefore $\ \left(\lambda_1-\lambda_2\right) y_2^\dagger x_1=0\ $, and, since $\ \lambda_2\ne\lambda_1\ $, it follows that $\ y_2^\dagger x_1=0\ $.  That is, $\ x_1\ $ is orthogonal to $\ y_2\ $. Thus, an eigenvector of $\ A\ $ corresponding to an eigenvalue $\ \lambda\ $ is orthogonal to every eigenvector of $\ A^\dagger\ $ corresponding to any eigenvalue $\ \mu\ne\lambda^*\ $, and an eigenvector of $\ A^\dagger\ $ corresponding to an eigenvalue $\ \mu\ $ is orthogonal to every eigenvector of $\ A\ $ corresponding to any eigenvalue $\ \lambda\ne\mu^*\ $.
If $\ AA^\dagger x = \lambda x\ $, with $\ x\ne 0\ $, then
\begin{align}
\|A^\dagger x\|^2&= x^\dagger AA^\dagger x\\
&=x^\dagger\left(\lambda x\right)\\
&=\lambda\|x\|^2\ .
\end{align}
Therefore,
$$
\lambda=\frac{\|A^\dagger x\|^2}{\|x\|^2}\ge 0
$$
