2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $X$. $\endgroup$ – Robert Lewis Nov 14 '19 at 16:28
  • $\begingroup$ You are right. Sorry about that. $\endgroup$ – mithusengupta123 Nov 14 '19 at 16:32
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.