TITLE: Minimum eigenvalue and singular value of a square matrix
QUESTION [7 upvotes]: How to show that the relationship $\left | \lambda_{min} \right | \geq \sigma_{min}$ holds between the minimum eigenvalue and singular value of a square matrix $A \in \mathbb{C}^{n \times n}$?

REPLY [10 votes]: On one hand, if $v$ is a unit eigenvector of $A$ corresponding to the eigenvalue $\lambda_\min$, then $v^\ast A^\ast Av=\|Av\|^2=|\lambda_\min|^2$. On the other hand, $A^\ast A$ is positive semidefinite. Hence it can be unitarily diagonalised as $U^\ast DU$, where $D$ is a nonnegative diagonal matrix. The minimum diagonal entry of $D$, i.e. the minimum eigenvalue of $A^\ast A$, is equal to $\sigma_\min^2$. Therefore
$$
|\lambda_\min|^2
=v^\ast A^\ast Av
\ge\min_{\|x\|=1}x^\ast A^\ast Ax
=\min_{\|x\|=1}(Ux)^\ast D (Ux)
=\min_{\|y\|=1}y^\ast D y
=\sigma_\min^2.
$$