For any square matrix A, we have

$-\sigma_{n-k+1}(A) \le \lambda_k\left(\frac{A+A^T}{2}\right) \le \sigma_k(A)$

where $0\le\sigma_n\le\dots\le\sigma_1$ are the singular values, and $\lambda_n\le\dots\le \lambda_1$ are the eigenvalues.

Proof of the right half of the inequality:

See Topics in matrix analysis: page 151, Corollary 3.1.5

Proof of the left half of the inequality:

First, we have

$\lambda_k\left(\frac{-A-A^T}{2}\right)=-\lambda_{n-k+1}\left(\frac{A+A^T}{2}\right)$

To verify, it is easy to see $\lambda_1\left(\frac{-A-A^T}{2}\right)=-\lambda_{n}\left(\frac{A+A^T}{2}\right)$ and $\lambda_n\left(\frac{-A-A^T}{2}\right)=-\lambda_{1}\left(\frac{A+A^T}{2}\right)$.

Second, applying the right half of the inequality gives

$\lambda_k\left(\frac{A+A^T}{2}\right)=-\lambda_{n-k+1}\left(\frac{-A-A^T}{2}\right)\ge-\sigma_{n-k+1}(-A)=-\sigma_{n-k+1}(A)$