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?


1 Answer 1


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 $\lambda \geq 0$.

  • $\begingroup$ Ah, I didn't realize it would be so simple.. Thank you very much! $\endgroup$ Jan 2, 2016 at 5:03

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