# Minimum eigenvalue and singular value of a square matrix

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 $A \in \mathbb{C}^{n \times n}$?

• I think it is worth mentioning that you implicitly order the eigenvalues by their modulus. – Julien May 4 '13 at 3:18

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 $$|\lambda_\min|^2 =v^\ast A^\ast Av \ge\min_{\|x\|=1}x^\ast A^\ast Ax =\min_{\|x\|=1}(Ux)^\ast D (Ux) =\min_{\|y\|=1}y^\ast D y =\sigma_\min^2.$$
This is related to the so called weyl-horn theorem on the relation between eigenvalues and singular values of an arbitrary matrix $A$. Let $\lambda_1,\lambda_2,\dots,\lambda_n$ be the eigenvalues, such that $|\lambda_1|>|\lambda_2|>\dots>|\lambda_n|$ (your $\lambda_{min}=\lambda_n$) and $\sigma_1,\sigma_2,...,\sigma_n$ (your $\sigma_{min}=\sigma_n$) be the singular values. Then \begin{align} &\prod_{j=1}^{k}|\lambda_j|\leq \prod_{j=1}^{k}\sigma_j~~,k=1,2,...,n-1 \\ &\prod_{j=1}^{n}|\lambda_j|= \prod_{j=1}^{n}\sigma_j \\ \end{align} Let $a=\prod_{j=1}^{n-1}|\lambda_j|$ and $b=\prod_{j=1}^{n-1}\sigma_j$, then we have $a \leq b$ and $a|\lambda_{min}|=b\sigma_{min}$ , thus $|\lambda_{min}|\geq \sigma_{min}$
• These are called Weyl's inequalities. What Horn did is that he proved that such sequences always occur as eigenvalues and singular values of some complex $n\times n$ matrix. Note: your argument requires to treat the null case separately. This being said, +1. – Julien May 4 '13 at 3:13