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Rayleigh quotient R(M,x), is defined as: $R(M,x) := {x^{*} M x \over x^{*} x}$

And it is said to be always less than the Largest eigenvalue: $R(M, x) \leq \lambda_\max$

Consider a vector $y$ with all real positive entries and normalized such that $y^Ty=1$

Can I say $My \leq \lambda_{max}y$ when $M$ is symmetric? (by transposing both side and multiply both side by $y$)

And Does $My \leq \lambda_{max}y$ means if $z=My$, $z_i \leq \lambda_{max} y_i$ for all $i$?

p.s. The system does not allow me to create the "rayleigh-quotient" tag :(

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No: take $ A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$ If $x = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $ then one of the entries of $Ax$ is $5$. On the other hand, the characteristic equation is $(\lambda-1)(\lambda-3)-4=0$, which has solutions $ 2 \pm \sqrt{5}$. Since $2 + \sqrt{5}<5$, we can conclude that its not always true that $Ax \leq \lambda_{\rm max} x$. Finally, you can divide $x$ by its norm (i.e., $\sqrt{2}$) to make sure $x^T x=1$ as you want.

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    @robinson oh I see the point. Thanks!!2011-09-09