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Eigenvalues of Symmetric Matrices

Linear Algebra
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Ram_1729
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Eigenvalues of Symmetric Matrices

#1

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 $B$
Demetres
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Re: Eigenvalues of Symmetric Matrices

#2

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.

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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