Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :(

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Thanks for your example!! However can anyone point out what is the algebraic mistake of the inequality? –  Benjamin Sep 9 '11 at 5:16
    
@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
    
@robinson oh I see the point. Thanks!! –  Benjamin Sep 9 '11 at 5:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.