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Thread: Symmetric Matrix proof !

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Avi06 View Post
    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, $\displaystyle |x|= |y|$.
    Last edited by HallsofIvy; Dec 11th 2009 at 08:34 AM.
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