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Math Help - Symmetric Matrix proof !

  1. #1
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    Symmetric Matrix proof !

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

    Show that:
    <br />
\lambda_{1}\|x\|^{2} \geq x^{T}Ax \geq \lambda_{n}\|x\|^{2}<br />

    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  \lambda_{1} \geq \lambda_{2} \geq .... \geq \lambda_{n} , where  \lambda_{i} are eigenvalues of A.

    Show that:
    <br />
\lambda_{1}\|x\|^{2} \geq x^{T}Ax \geq \lambda_{n}\|x\|^{2}<br />

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

    Finally, since P is orthogonal, |x|= |y|.
    Last edited by HallsofIvy; December 11th 2009 at 08:34 AM.
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