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.

  • $\begingroup$ Why is it true if $A$ is symmetric? $\endgroup$ – Oria Gruber Feb 3 '14 at 16:02
  • $\begingroup$ If $A$ is symmetric, then it is diagonalizable by an orthonormal matrix $A=U^TDU$, and thus $\max_{\|u\|=1} \|Au\|=\max_{\|u\|=1}\langle u,Au\rangle=\max_{\lambda\in \sigma(A)}\lambda$. $\endgroup$ – Yiorgos S. Smyrlis Feb 3 '14 at 16:05
  • $\begingroup$ 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? $\endgroup$ – Oria Gruber Feb 3 '14 at 16:13
  • $\begingroup$ $\max\langle u,Au\rangle=\max\langle Uu,DUu\rangle=\max\langle v,Dv\rangle=\max \lambda$ $\endgroup$ – Yiorgos S. Smyrlis Feb 3 '14 at 16:18
  • 1
    $\begingroup$ 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. $\endgroup$ – Yiorgos S. Smyrlis Feb 3 '14 at 16:23

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