Spectral theorem: eigenvalue decomposition for symmetric matrices
Proof:
The proof is by induction on the size of the matrix . The result is trivial for . Now let and assume the result is true for any matrix of size .
Consider the function of , . From the basic properties of the determinant, it is a polynomial of degree , called the characteristic polynomial of . By the fundamental theorem of algebra, any polynomial of degree has (possibly not distinct) complex roots; these are the called the eigenvalues of . We denote these eigenvalues by .
If is an eigenvalue of , that is, , then must be non-invertible (see here). This means that there exist a non-zero real vector such that . We can always normalize so that . Thus, is real. That is, the eigenvalues of a symmetric matrix are always real.
Now consider the eigenvalue and an associated eigenvector . Using the Gram-Schmidt orthogonalization procedure, we can compute a matrix such that is orthogonal. By induction, we can write the symmetric matrix as , where is a matrix of eigenvectors, and are the eigenvalues of . Finally, we define the matrix . By construction the matrix is orthogonal.
We have
where we have exploited the fact that , and .
We have exhibited an orthogonal matrix such that is diagonal. This proves the theorem.
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