# Linear algebra - question about vector norm and eigenvalues

Maybe a basic question, but I'd like to know the reasoning behind it if its true.

suppose I have a matrix $A \in \mathrm{Mat}_n(\mathbb R)$ with the eigenvalues $\lambda_1 ,\lambda_2 ,..., \lambda_k$.

Suppose $\forall j\neq i, |\lambda_i|\geq|\lambda_j|$ (Meaning $\lambda_i$ is the eigenvalue that is largest absolute value).

Suppose $v_i$ is a respective eigenvector of $\lambda_i$ such that $Av_i=\lambda_iv_i$ and $||v_i||=1$.

is it true that $||Av_i|| \geq ||Au||$ when $u$ is some vector in $\mathbb R^n$ such that $||u||=1$?

The answer is no $$A=\left(\begin{matrix}1&1 \\ 0&1\end{matrix}\right).$$ This matrix has only one eigenvalue $\lambda=1$, and its eigenvector is $u=(1,0)$. Take $$v=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right),$$ then $\|v\|=1$ and $$Av=\left(\sqrt{2},\frac{\sqrt{2}}{2}\right),$$ with $\|Av\|=\dfrac{\sqrt{10}}{2}>1=\|u\|$.
However, what you are saying is TRUE if $A$ is symmetric.
• Why is it true if $A$ is symmetric? – Oria Gruber Feb 3 '14 at 16:02
• If $A$ is symmetric, then it is diagonalizable by an orthonormal matrix $A=U^TDU$, and thus $\max_{\|u\|=1} \|Au\|=\max_{\|u\|=1}\langle u,Au\rangle=\max_{\lambda\in \sigma(A)}\lambda$. – Yiorgos S. Smyrlis Feb 3 '14 at 16:05
• It is unclear to me why $max<u,Au>=max \lambda$. I understand it's true and it is intuitevly correct, but how do I show it? – Oria Gruber Feb 3 '14 at 16:13
• $\max\langle u,Au\rangle=\max\langle Uu,DUu\rangle=\max\langle v,Dv\rangle=\max \lambda$ – Yiorgos S. Smyrlis Feb 3 '14 at 16:18
• What I mean is the following: If $u$ runs all the unit vectors, then so does $v=Uu$, as $U$ is orthogonal, and so is its inverse. – Yiorgos S. Smyrlis Feb 3 '14 at 16:23