Let A be a real square matrix. How to show that $\lambda_{min}((A^TA)^{\frac{1}{2}})\geq \lambda_{min}( (A+A^T)/2)$? Here, $\lambda_{min}()$ denotes the minimal eigenvalue of a Hermitian matrix.
2026-04-08 15:49:38.1775663378
how to prove the smallest singular value of a matrix is no smaller than the minimal eigenvalue of its symmetric part?
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