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 $\lambda \in \lambda_1, ..., \lambda_n$, \begin{equation} \|{Ax}\|_{2}\geq |\lambda| \; \|x\|_{2} \end{equation} 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 $A$: \begin{equation} A = \sum_{i=1}^{n} \lambda _{i} v_{i}v_{i}^\intercal \end{equation} 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 \begin{equation} \|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 \end{equation}
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..
