# Symmetric Matrix proof !

• Dec 11th 2009, 05:27 AM
Avi06
Symmetric Matrix proof !
Given that A is a n x n symmetric matrix and $\displaystyle \lambda_{1} \geq \lambda_{2} \geq .... \geq \lambda_{n}$ , where $\displaystyle \lambda_{i}$ are eigenvalues of A.

Show that:
$\displaystyle \lambda_{1}\|x\|^{2} \geq x^{T}Ax \geq \lambda_{n}\|x\|^{2}$

Having trouble proving the above ! Any help would be much appreciated.
• Dec 11th 2009, 08:11 AM
HallsofIvy
Quote:

Originally Posted by Avi06
Given that A is a n x n symmetric matrix and $\displaystyle \lambda_{1} \geq \lambda_{2} \geq .... \geq \lambda_{n}$ , where $\displaystyle \lambda_{i}$ are eigenvalues of A.

Show that:
$\displaystyle \lambda_{1}\|x\|^{2} \geq x^{T}Ax \geq \lambda_{n}\|x\|^{2}$

Having trouble proving the above ! Any help would be much appreciated.

Since A is symmetric, it is diagonalizable. That is, there exist orthogonal P such that $\displaystyle PAP^T= D$ where D is the diagonal matrix with the eigenvalues of A on the diagonal. In that case, $\displaystyle A= P^TDP$ so that $\displaystyle x^TAx= x^T(P^TDP)x= (x^TP^T)D(Px)= (Px)^TD(Px)$ or, letting y= Px, $\displaystyle x^TAx= y^TDy$. Now it is clear that $\displaystyle y^TDY= \sum \lambda_i y_i^2$ which is necessarily larger than $\displaystyle \lambda_1|y|^2= \sum \lambda_1 y_n^2$ and smaller than $\displaystyle \lambda_{n}|y|^2= \sum \lambda_{n}y_n^2$.

Finally, since P is orthogonal, $\displaystyle |x|= |y|$.