Eigenvalues and Singular Values of Negative Definite Matrix If $A$ is positive definite ($A^T=A$ and $\lambda_i(A)>0$) then the singular values $\sigma_i(A)$ coincide with the eigenvalues $\lambda_i(A)$. This fact was proved in another answer, and it made me curious as to whether the same (or something similar) happens when $A$ is negative definite $(A^T=A$ and $\lambda_i(A)<0$).
 A: If $A$ is negative definite, then $B=-A$ is definite positive and $BB^T = AA^T$. The singular values of $B$ are therefore equal to the one of $A$ and also to the eigenvalues of $A$ according to the result you mention.
While the eigenvalues of $B$ are the opposite of the one of $A$.
Finally, the singular values of $A$ are the opposite of its eigenvalues. This is just saying that $\sqrt{\vert \lambda \vert} = \sqrt{- \lambda}$ for $\lambda \le 0$.
A: Singular values are always nonnegative, but the eigenvalues of a negative definite matrix are always negative. So, the singular values cannot possibly be equal to the eigenvalues in this case.
However, in general, when $A$ is normal (this includes the case where $A$ is negative definite), its singular values are precisely the moduli of its eigenvalues, i.e. $\sigma_k(A)=|\lambda_(A)|$. Let $\lambda_1(A),\ldots,\lambda_n(A)$ be the eigenvalues of $A$ and let $\lambda_k(A)=|\lambda_k(A)|e^{i\theta_k}$ for each $k$. Since $A$ is normal, it can be unitarily diagonalised as $A=U\Lambda U^\ast$. Therefore $A=(UD)|\Lambda|U^\ast$ is a singular value decomposition of $A$, where $|\Lambda|=\operatorname{diag}(|\lambda_1(A)|,\ldots,|\lambda_n(A)|)$ and $D=\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})$. It follows that the singular values of $A$ are $|\lambda_1(A)|,\ldots,|\lambda_n(A)|$.
A: By definition $\sigma_i(A)=\sqrt{\lambda_i(A^TA)}$. If $A=A^T$ then $\sigma_i(A)=\sqrt{\lambda_i(A^2)}$. By eigendecomposition $A=U\Sigma U^T$ we see that $A^2=U\Sigma^2 U^T$ so $\lambda_i(A^2)=\lambda_i(A)^2$. We conclude that $\sigma_i(A)=\sqrt{\lambda_i(A)^2}=|\lambda_i(A)|$. This is true for symmetric matrices in general, we do not need to assume $A$ is positive or negative definite.
