Estimating the shift in the $\lambda_{max}$ of a matrix under a diagonal perturbation

Question

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 $1,-1,0$)

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I see some related questions like,

• Spectral radius of a rank-1 perturbation

• Relation between the subordinate norm and the spectral radius of a matrix

• The spectral radius of a modified graph

• Eigenvectors as continuous functions of matrix - diagonal perturbations

Answer 1

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.

Answer 2

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*}