A theorem on positive semidefinite forms and eigenvalues

Alicia Y. Tsai; Authors: Laurent EI Ghaoui; Contributors: Phan Hoang, Tony Tin, Ha To Tam, Nguyen Van Kep, Le Dinh Nam, Juliette Decugis, Nguyen Quoc Cuong.; Giuseppe C. Calafiore

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 $x$ :

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

Conversely, if there exist $i$ for which $\lambda_i <0$ , then choosing $x = u_i$ will result in $q(u_i) 0$ for every $i$ , then the condition

$\begin{align*} q_A(x) &= x^TAx = \sum\limits_{i=1}^n \lambda_i (u_i^T x)^2 = 0 \end{align*}\end{align*}$

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 $x = u_i$ .

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Theorem: (Link with SED)

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