# A question about Rayleigh quotient

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.
@Benjamin - the mistake is that a false statement can imply a true one. For example, the false statement $2=-2$ upon squaring implies the true statement $4=4$. In your case, the false statement $My \leq \lambda_{\rm max} y$ implies, upon multiplying from the left by $y^T$, the true statement $y^T M y \leq \lambda_{\rm max} y^T y$. – robinson Sep 9 '11 at 5:19
@robinson how about I do it in that way: $y^TMy\leq\lambda_{max}$ $yMy^T \leq \lambda_{max}$ (I am not sure I can do this or not, transposing both side) $yM \leq \lambda_{max}y$ – Benjamin Sep 9 '11 at 5:26
@Benjamin - when you transpose $y^T M y$, you get $y^T M y$. Indeed, if $y$ is a column vector, then $yMy^T$ does not even make sense - how do you multiply a column vector on the right by a matrix? – robinson Sep 9 '11 at 5:28