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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?

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  • $\begingroup$ 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 $X$. $\endgroup$ Commented Nov 14, 2019 at 16:28
  • $\begingroup$ You are right. Sorry about that. $\endgroup$ Commented Nov 14, 2019 at 16:32

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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, $$ \lambda=\frac{\|A^\dagger x\|^2}{\|x\|^2}\ge 0 $$

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