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Given a matrix $A$ and a diagonal matrix $D$, what ways do we have to estimate, $\lambda_{max}(A+D) - \lambda_{max}(A)$? (Feel free to make other assumptions about the matrices that they are all symmetric and have entries in $1,-1,0$)


I see some related questions like,

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2 Answers 2

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Because $A \mapsto \|A\|_2 := \underset{x \ne 0}{\text{sup }}\frac{\|Ax\|_2}{\|x\|_2}$ defines a matrix norm, we have the triangle inequality $\|A + D\|_2 \le \|A\|_2 + \|D\|_2$, i.e \begin{eqnarray} \|A + D\|_2 - \|A\|_2 \le \|D\|_2 = \underset{1 \le i \le n}{\text{sup }}|d_i|. \end{eqnarray}

Further because the spectral radius $r(A) := sup\{|\lambda|| \lambda \in spect(A)\}$ is upper-bounded by $\|A\|_2$ (exercise), we have \begin{eqnarray} r(A) - \|A\|_2 \le \underset{1 \le i \le n}{\text{sup }}|d_i|. \end{eqnarray}

I don't think you can get anything tighter without further assumptions.

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  • $\begingroup$ Can you kindly explain how you dissolved $\vert \vert A + D \vert \vert_2$ ? $\endgroup$ Aug 30, 2015 at 21:25
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    $\begingroup$ $\|A\|_2$ is not the right quantity to look at. Just consider the case $\lambda_{\max}(A)\leq0$ (i.e., $A\leq0$) in which $\|A\|_2=-\lambda_{\min}(A)$ is as far off from $\lambda_{\max}(A)$ as possible. Besides, $r(A)=\|A\|_2$ for a symmetric matrix $A$. $\endgroup$
    – ifw
    Aug 30, 2015 at 23:24
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You need $A$, $D$ symmetric to guarantee that the eigenvalues are real. Then you have the elementary estimate \begin{multline*} \max\{\lambda_{\max}(A)+\lambda_{\min}(D),\lambda_{\min}(A)+\lambda_{\max}(D)\} \\ \leq \lambda_{\max}(A+D) \leq \lambda_{\max}(A)+\lambda_{\max}(D), \end{multline*} and this estimate is sharp as easy examples with $A$ being diagonal show.

EDIT: For instance, \begin{multline*} \lambda_{\max}(A+D) = \max_{\|x\|_2=1}\langle (A+D)x,x\rangle \\ \geq \max_{\|x\|_2=1}\langle Ax,x\rangle + \min_{\|x\|_2=1}\langle Dx,x\rangle =\lambda_{\max}(A) + \lambda_{\min}(D). \end{multline*}

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  • $\begingroup$ This is a good answer. You have my upvote. But you should add a reference to Weyl inequalities (which are far from elementary, given the state of affairs...) or other combo. Statements like "this or that is elementary" are not very useful to the user. $\endgroup$
    – dohmatob
    Aug 31, 2015 at 6:43
  • $\begingroup$ @dohmatob This particular case of the Weyl inequalities is indeed straightforward. I've made an edit to my answer. $\endgroup$
    – ifw
    Aug 31, 2015 at 7:02

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