A theorem on positive semidefinite forms and eigenvalues

L. El Ghaoui

A theorem on positive semidefinite forms and eigenvalues

Theorem: (Link with SED)

A quadratic form $q(x) = x^TAx$ , with $A \in {\bf S}^n$ 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 $x$ :

$q_A(x) = x^TAx = \sum\limits_{i=1}^n \lambda_i (u_i^T x)^2 \ge 0.$

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

$q_A(x) = x^TAx = \sum\limits_{x=1}^n \lambda_i (u_i^T x)^2 = 0$

trivially implies $u_i^T x$ 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 $\lambda_i \leq 0$ for some $i$ , we can achieve $q(x) \leq 0$ for some non-zero $x = u_i$ .

A theorem on positive semidefinite forms and eigenvalues

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