Theorem: (Link with SED)
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A quadratic form [latex]q(x) = x^TAx[/latex], with [latex]A \in {\bf S}^n[/latex] is non-negative (resp. positive-definite) if and only if every eigenvalue of the symmetric matrix [latex]A[/latex] is non-negative (resp. positive). |
Proof: Let [latex]A = U^T\Lambda U[/latex] be the SED of [latex]A[/latex].
If [latex]A \succeq 0[/latex], then [latex]\lambda_i \ge 0[/latex] gor every [latex]i[/latex]. Thus, for every [latex]x[/latex]:
[latex]\begin{align*} q_A(x) &= x^TAx = \sum\limits_{i=1}^n \lambda_i (u_i^T x)^2 \ge 0. \end{align*}[/latex]
Conversely, if there exist [latex]i[/latex] for which [latex]\lambda_i <0[/latex], then choosing [latex]x = u_i[/latex] will result in [latex]q(u_i) 0[/latex] for every [latex]i[/latex], then the condition
[latex]\begin{align*} q_A(x) &= x^TAx = \sum\limits_{i=1}^n \lambda_i (u_i^T x)^2 = 0 \end{align*}\end{align*}[/latex]
trivially implies [latex]u_i^T x[/latex] for every [latex]i[/latex], which can be written as [latex]Ux=0[/latex].
Since [latex]U[/latex] is orthogonal, it is invertible, and we conclude that [latex]x=0[/latex]. Conversely, if [latex]\lambda_i \leq 0[/latex] for some [latex]i[/latex], we can achieve [latex]q(x) \leq 0[/latex] for some non-zero [latex]x = u_i[/latex].
