For positive semi-definite matrices, each singular value is equal to eigenvalue

Question

It is known that If a matrix is PSD (symmetric), then its eigenvalues are equal to its singular value.

How to prove it? Hope for a hint.

thanks

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Thanks for the answer from Three. OneFour; the following is the potential answer:

Let $A = U\Lambda U^T$, we know $AA^T=U\Lambda^2U^T$. By the definition of $\sigma(A)=(\lambda(AA^T))^{\frac{1}{2}}$ and the fact that all eigenvalues are positive, we have $$\sigma(A)=(\Lambda^2)^{\frac{1}{2}}=\lambda(A)$$

Note that,

1. If $A$ is just symmetric, then $\lambda_+(A)$ is equal to the corresponding $\sigma(A)$.
2. For general square matrices $A$, we have the conclusion $\lambda(A)\leq \sigma(A)$.

I hope my answer makes sense.

Answer 1

Let $A$ be a symmetric, positive definite matrix with singular value decomposition $A = U \Sigma V^T$ (i.e. $U$, $V$ are orthogonal matrices and $\Sigma$ a diagonal matrix). Now look at how the singular values of $A$ are related to the eigenvalues of $A^T A$. Then think about how the eigenvalues of $A$ are related to the eigenvalues of $A^T A$ and use that if $A$ is symmetric, positive definite then its eigenvalues are real and positive.