A theorem on positive semidefinite forms and eigenvalues

Theorem: (Link with SED)

A quadratic form q(x) = x^TAx, with A in mathbf{S}^n is non-negative (resp. positive-definite) if and only if every eigenvalue of the symmetric matrix A is non-negative (resp. positive).

Proof: Let A = U^T Lambda U be the SED of A.

If A succeq 0, then lambda_i ge 0 gor every i. Thus, for every x:
 q_A(x) = x^TAx = sum_{i=1}^n lambda_i (u_i^Tx)^2 ge 0.
Conversely, if there exist i for which lambda_i<0, then choosing x=u_i will result in q(u_i) <0.

Likewise, a matrix A is PD if and only if q_A is a positive-definite function, that is, q_A(x) ge 0 for every x, and q_A(x) = 0 if and only if x = 0. When lambda_i >0 for every i, then the condition
 q_A(x) = x^TAx = sum_{i=1}^n lambda_i (u_i^Tx)^2 = 0
trivially implies u_i^Tx = 0 for every i, which can be written as Ux = 0. Since U is orthogonal, it is invertible, and we conclude that x = 0. Conversely, if lambda_i le 0 for some i, we can achieve q(x) le 0 for some non-zero x = u_i.

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