Exercise 6.1: Bounds on Eigenvalues

chapter 6

Let $A$ and $B$ be two symmetric $n\times n$ matrices. Let $u$ be the eigenvector corresponding to the largest eigenvalue of $A$, and let $v$ be the eigenvector corresponding to the largest eigenvalue of $B$. Then $-\lra{v, B v} \le -\lra{u, B u}$, so \begin{equation} \lra{u, A u} - \lra{v, B v} \le \lra{u, (A - B) u} \le \norm{u}_2 \norm{(A - B) u}_2 \le \norm{A - B}_2. \end{equation} Therefore, \begin{equation} \abs{\lambda_1(A) - \lambda_1(B)} \le \norm{A - B}_2, \end{equation} and we similarly prove that \begin{equation} \abs{\lambda_n(A) - \lambda_n(B)} \le \norm{A - B}_2. \end{equation}