Semidefiniteness of Quadratic form with zero determinant

Question

Consider a function in quadratic form $f=x^T A x$, where $x\in\mathbb R^2$.Let $\det A=0$. (Therefore, one eigenvalue is zero. )

With the above assumption, how can it be shown that $f: x\mapsto f$ is $\geq0$ or $\leq 0$ for all $x$?

I assume that if $\lambda_1=0, \lambda_2>0$, $f\geq0$ and vice versa. But not sure.

Answer 1

Note that since $f$ is a quadratic form, $A$ is symmetric. Assume $\lambda_1 \le \lambda_2$. By the min-max theorem, we have

$$ \lambda_1 \le \frac{x^TAx}{x^Tx} \le \lambda_2\tag1 $$

for all non-zero $x$. Note that $x^Tx \ge 0$. There are two cases:

• $\lambda_1=0$ and $\lambda_2 \ge 0$,
• $\lambda_1 < 0$ and $\lambda_2 =0$.

We see from $(1)$ that in the first case $f(x)=x^TAx$ is non-negative for all $x$. Similarly, in the second case $f(x)$ is non-positive for all $x$.

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The statement is not true if $A$ is not symmetric. Let

$$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}\quad u = \begin{pmatrix}1 \\ -1\end{pmatrix}\quad v = \begin{pmatrix}1 \\ 1\end{pmatrix}. $$

Then

$$ f(u) = u^TAu = -1 \\ f(v) = v^TAv = 1. $$

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Also, the statement is not true in $\mathbb{R}^n$ for $n>2$ even if $A$ is symmetric. Counterexample: $A=\mathrm{diag}(1, -1, 0)$.