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In a physics book that I have, the author states a very important result but without proving it. He does provide clues for the proof though! It goes like this..

Consider the eigenvalue equation

[tex](M^{-1}K-\omega ^2 I)\vec{v} = \vec{0}[/tex]

where M is an n x n diagonal matrix

Edit: The elements of the K matrix are all positive too!

Consider the eigenvalue equation

[tex](M^{-1}K-\omega ^2 I)\vec{v} = \vec{0}[/tex]

where M is an n x n diagonal matrix

**whose elements are all positive**(first clue) and K is a**symetric**(second clue) n x n matrix. Then the matrix [itex]M^{-1}K[/itex] has exactly n linearly independant eigenvectors.Edit: The elements of the K matrix are all positive too!

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