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

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

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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.

[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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