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Why does one assume that the eigenbasis for a quadratic form is orthogonal, hence orthogonal diagonalization. I understand that for hermitian and unitary maps one can show by spectral theorem an orthogonal basis of eigenvectors exists, however does this also hold true for quadratic forms? My question is basically how does one show that a quadratic form has an orthogonal basis that furthermore also leads to a diagonal matrix? Thanks in advance for any insight

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  • $\begingroup$ In short, a quadratic form can be represented by a hermitian matrix. $\endgroup$ – amd Apr 13 at 23:09
  • $\begingroup$ Oh yes, I see the transpose of a matrix's complex conjugate is defined as the adjoint of the matrix. Therefore if one can represent the quadratic form as a matrix that is its own adjoint, we are done? $\endgroup$ – Rick Apr 13 at 23:13
  • $\begingroup$ Is this why a quadratic form is constrained to a symmetric bilinear form $\endgroup$ – Rick Apr 13 at 23:17
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For a symmetric matrix the eigenvectors associated with distinct eigenvalues are orthogonal.

The eigenvectors associated to the same eigenvalues are orthogonalized by the Gram-Schmidt process.

Thus for a quadratic form we always have an orthogonal set of eigenvectors.

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