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I'm trying to show that if A is an Hermitian matrix with non-negative eigenvalues, then A is positive semi-definite.

The only thing I've thought of so far is using Spectral Theorem. I know I want to show that $x^*Ax\ge0, \forall x\in\mathbb{C}^n$, therefore I need to show the following (the ... part):

$x^*Ax = x^*U^*DUx = ... \ge0$, where $U$ is unitary, $D$ is diagonal with non-negative entries.

Thank you for any assistance.

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  • $\begingroup$ Write $D=E^\star E$, where $E$ is diagonal, then you have $(EUx)^\star(EUx)$. $\endgroup$ – vadim123 Nov 29 '14 at 2:36
  • $\begingroup$ Surely this is a duplicate. $\endgroup$ – user_of_math Nov 29 '14 at 3:12
  • $\begingroup$ I was unable to find a similar question. $\endgroup$ – John Nov 29 '14 at 3:13
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Please let me know if there is an issue with this answer:

Let $D = E^*E$ where E is some diagonal matrix.

$x^*Ax = x^*U^*DUx = x^*U^*E^*EUx = (EUx)^*(EUx) = \left< EUx,EUx\right>\ge0$. The final step is due to properties of inner products of complex vectors.

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