# 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?

• I don't see how you get from $X^\dagger(A^\dagger Y - \lambda^\ast Y) = 0$ to $A^\dagger Y = \lambda^\ast Y$; all I can get is that $A^\dagger Y - \lambda^\ast Y$ is in the null space of the linear map functional $Z \to X^\dagger Z$, i.e. is normal (in the sense of the hermitian inner product $X^\dagger Y$) to $X$. – Robert Lewis Nov 14 '19 at 16:28
• You are right. Sorry about that. – mithusengupta123 Nov 14 '19 at 16:32

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$$