Welcome to mathimatikoi.org forum; Enjoy your visit here.

Eigenvalues of Symmetric Matrices

Linear Algebra
Post Reply
Articles: 0
Posts: 4
Joined: Mon Dec 03, 2018 4:59 pm

Eigenvalues of Symmetric Matrices


Post by Ram_1729 » Mon Dec 03, 2018 5:44 pm

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 $B$
Former Team Member
Articles: 0
Posts: 77
Joined: Mon Nov 09, 2015 11:52 am
Location: Limassol/Pyla Cyprus

Re: Eigenvalues of Symmetric Matrices


Post by Demetres » Mon Dec 03, 2018 10:12 pm

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.


\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\|

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
Post Reply