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 Post subject: Eigenvalues of Symmetric MatricesPosted: Mon Dec 03, 2018 5:44 pm

Joined: Mon Dec 03, 2018 4:59 pm
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For any two symmetric $n\times n$ matrices $A$ and $B$ be their eigenvalues be ordered from largest to smallest . How to prove that for eigenvalues $|\lambda _k^A-\lambda _k^B|\le ||A-B||$ for $1\le k\le n$. Where $\lambda _k^A,\lambda _k^B$ are respective eigenvalues of $A$ and $B$

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 Post subject: Re: Eigenvalues of Symmetric MatricesPosted: Mon Dec 03, 2018 10:12 pm

Joined: Mon Nov 09, 2015 11:52 am
Posts: 77
Location: Limassol/Pyla Cyprus
This follows from the following characterisation of the eigenvalues of symmetric matrices.

$\lambda_k(A) = \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^TAx}{\|x\|_2}$

Here the minimum is taken over all subspaces $U$ of $\mathbb{R}^n$ of dimension $n-k+1$. (I chose to write $\lambda_K(A)$ rather than $\lambda_k^A$.) The proof of this uses the fact that symmetric matrices have an orthonormal basis of eigenvectors.

Now,

\begin{aligned} \lambda_k(A) &= \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^TAx}{\|x\|_2} \\ &= \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^T(A-B+B)x}{\|x\|_2} \\ &= \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \left[ \frac{x^TBx}{\|x\|_2} + \frac{x^T(A-B)x}{\|x\|_2}\right] \\ &\leqslant \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \left[ \frac{x^TBx}{\|x\|_2} + \|A-B\|\right] \\ &= \lambda_k(B) + \|A-B\| \end{aligned}

So $\lambda_k(A) - \lambda_K(B) \leqslant \|A-B\|$. Analogously we have $\lambda_k(B) - \lambda_K(A) \leqslant \|A-B\|$

This inequality is a particular case of Weyl's Inequalities

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