Why does completing the square force every self-adjoint operator on a real space to have an eigenvalue — and then an orthonormal eigenbasis?
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invertible when
For self-adjoint and nonzero , expand , bound the middle term by Cauchy–Schwarz, complete the square to show it is positive, and conclude injectivity, hence invertibility.
Self-adjoint minimal polynomial: a product of real linear factors
Treat through real eigenvalues; over , factor the minimal polynomial and use the invertible quadratics to rule out every quadratic factor by minimality of degree, giving an eigenvalue.
The three equivalent conditions and their proof
State (a) self-adjoint, (b) diagonal in some orthonormal basis, (c) orthonormal eigenbasis; prove (a)⇒(b) from an orthonormal upper-triangular matrix equal to its transpose, then (b)⇒(a) and (b)⇔(c).
Worked example: a symmetric matrix
For the matrix with rows , , , verify the eigenvectors , , , normalize them, and record the diagonal matrix .