Eigenvectors of symmetric matrices

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  • #1
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Can anyone prove that the eigenvectors of symmetric matrices are orthogonal?
 

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  • #2
HallsofIvy
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Let v be a eigenvector with eigenvalue [itex]\lambda_1[/itex] and u an eigenvector with eigenvalue \(\displaystyle \lambda_2\), both with length 1.

[itex]\lambda_1<v, u>= <\lambda_1v, u>[/itex]
(<u, v> is the innerproduct)
[itex]= < Av, u>= \overline{<v, Au>}[/itex]
(because A is symmetric)
("self adjoint" in general)
[itex]= \overline{<v, \lambda_2u>}= \lambda<v, u>[/itex]
so that
[itex]\lambda_1<v, u>= \lambda_2<v, u>[/itex]
[itex](\lambda_1- \lambda_2)<v, u>= 0[/itex]
Since [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] are not equal,
[itex]\lambda_1- \lambda_2[/itex] is not 0, <v, u> is.
 
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  • #3
mathman
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The important criterion is not symmetry, but whether the eigenvalues are all different (i.e. all the roots of the characteristic equation are different). If an eigenvalue is repeated, then the space associated with the eigenvector is not one dimensional, so that the vectors are not necessarily orthogonal.
 

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