2
$\begingroup$

Suppose $T^*$ is the adjoint of $T.$ I want to prove that $T^*T$ has only non-negative real eigenvalues.

I proved that $T^*T$ is self-adjoint (because $\overline{(\overline{A^T}A)^T} = \overline{A^T}A,$ where $A^T$ is the transpose and $\overline{A}$ is its conjugate transpose and $A$ is the matrix of $T.$ Therefore, I can conclude that the eigenvalues of $T^*T$ are real. However, I don't know how to show they are all non-negative. I suppose showing $T^*T$ is positive semi-definite would do the job, but how can I show this? Or is there another way?

$\endgroup$
7
$\begingroup$

Let $\lambda \in \mathbb{R}$ be an eigenvalue of $T^* T$ and $v \neq 0$ an associated eigenvector. Then

$\lambda \langle v, v\rangle = \langle T^* T v, v\rangle = \langle Tv, Tv\rangle \geq 0$

implies $\lambda \geq 0$.

$\endgroup$
  • $\begingroup$ Ah, I didn't realize it would be so simple.. Thank you very much! $\endgroup$ – dhk628 Jan 2 '16 at 5:03

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.