# Matrix norm inequality involving smallest eigenvalue reference request

I search a reference (textbook/paper) for the following symmetric matrix norm inequality.

Let $$A \in \mathbb{R}^{n \times n}$$ be a real symmetric matrix and let $$\lambda_1, ..., \lambda_n \in \mathbb{R}$$ be its eigenvalues. We have that for all $$x \in \mathbb{R}^{n}$$ and for all $$\lambda \in \lambda_1, ..., \lambda_n$$, $$$$\|{Ax}\|_{2}\geq |\lambda| \; \|x\|_{2}$$$$ That is, denoting the smallest eigenvalue as $$\lambda _{s} = \min_{\lambda \in \left\{ \lambda_1, ..., \lambda_n\right\} } |\lambda|$$, $$\|Ax\|_{2} \geq \lambda _{s} \|x\|_{2}$$.

I can prove this inequality, but I need a textbook/paper proof for reference.. Do you happen to know any?

For completeness, my example verbose proof is:

Since $$A$$ is symmetric, spectral theorem applies and there exists a unique orthonormal basis formed by eigenvectors $$v_{1}, \dots, v_{n}$$ of $$A$$. We obtain the spectral decomposition of $$A$$: $$$$A = \sum_{i=1}^{n} \lambda _{i} v_{i}v_{i}^\intercal$$$$ The outer products $$v_{i}v_{i}^\intercal$$ are the orthogonal projections onto one-dimensional $$\lambda _{i}$$-eigenspace. Note that \begin{align*} \|Ax\|^{2} & = \left( \sum_{i=1}^{n} \lambda _{i} v_{i}v_{i}^\intercal x \right)^\intercal \left( \sum_{i=1}^{n} \lambda _{i} v_{i}v_{i}^\intercal x \right) \\ &= \sum_{i=1}^{n} \left( \lambda _{i} x^\intercal v_{i} v_{i}^\intercal \left( \lambda _{i} v_{i} v_{i}^\intercal x + \sum_{\substack{j=1 \\ j \not = i}}^{n} \lambda _{j} v_{j} v_{j}^\intercal x \right) \right) \\ &= \sum_{i=1}^{n} \lambda _{i}^{2} x^\intercal v_{i} \underbrace{v_{i}^\intercal v_{i}}_{1} v_{i}^\intercal x + \sum_{\substack{j=1 \\ j \not = i}}^{n} \lambda _{i} \lambda _{j} x^\intercal v_{i} \underbrace{v_{i}^\intercal v_{j}}_{0} v_{j}^\intercal x\\ &= \sum_{i=1}^{n} \lambda _{i}^{2} ( x^\intercal v_{i} ) ( v_{i}^\intercal x ) = \sum_{i=1}^{n} \lambda _{i}^{2} ( v_{i}^\intercal x )^{2} \end{align*} Also, given that any $$x$$ can by expressed as $$\sum_{i=1}^{n} ( v_{i}^\intercal x ) v_{i}$$ in the orthonormal base formed by eigenvectors $$v_{1},\dots,v_{n}$$ (i.e. as vector projections of $$x$$ onto $$v_{i}$$), \begin{align*} \|x\|^{2} &= x^\intercal x = \left( \sum_{i=1}^{n} v_{i}( v_{i}^\intercal x ) \right) ^\intercal \left( \sum_{i=1}^{n} v_{i} ( v_{i}^\intercal x ) \right) \\ &= \sum_{i=1}^{n} \left( ( x^\intercal v_{i} ) v_{i}^\intercal \left( v_{i}( v_{i}^\intercal x ) + \sum_{\substack{j=1 \\ j \not =i}}^{n} v_{j} (v_{j}^\intercal x) \right) \right) \\ &= \sum_{i=1}^{n} ( x^\intercal v_{i} ) \underbrace{v_{i}^\intercal v_{i}}_{1} ( v_{i}^\intercal x ) + \sum_{\substack{j=1 \\ j \not =i}}^{n} ( x^\intercal v_{i} ) \underbrace{v_{i}^\intercal v_{j}}_{0} ( v_{j}^\intercal x ) \\ &= \sum_{i=1}^{n} ( x^\intercal v_{i} ) ( v_{i}^\intercal x ) = \sum_{i=1}^{n} ( v_{i}^\intercal x ) ^{2} \end{align*} Now, we get an inequality $$$$\|Ax\|^{2} = \sum_{i=1}^{n} \lambda _{i}^{2} ( v_{i}^\intercal x )^{2} \geq \min_{j \in \left\{ 1,..,n \right\} } \lambda _{j}^{2} \sum_{i=1}^{n} ( v_{i}^\intercal x ) ^{2} = \min_{j \in \left\{ 1,..,n \right\} } \lambda _{j}^{2} \|x\|^2$$$$

Which after taking the square root finishes the proof.

The reference could have generalized or concise version of this proof, I just can't find anything..

There are some unnecessary steps in your proof. By eliminating them you can simplify it somewhat. Let $$\ x_i=v_i^\intercal x\$$. Then $$\ x=\sum_\limits{i=1}^nx_iv_i\$$, and \begin{align} \|Ax\|^2&=\Big\|A\sum_{i=1}^nx_iv_i\Big\|^2\\ &=\Big\|\sum_{i=1}^n\lambda_i x_iv_i\Big\|^2\\ &=\sum_{i=1}^n\lambda_i^2x_i^2\\ &\ge\lambda_s^2\sum_{i=1}^nx_i^2\\ &=\lambda_s^2\|x\|^2 \end{align}