Generalizing $x^\intercal A x = 0$ theorems to $x^\intercal A(x) x = 0$ forms?

Question

Let $x \in \mathbb{R}^n$ be any real vector and let$\ A = A^\intercal \in \mathbb{R}^{n \times n}\ $ be a real symmetric matrix. Then, $$ x^\intercal A x = 0\ \ \ \forall x \ \ \implies \ \ A \equiv 0 $$

Can we still say this if $\ A = A^\intercal : \mathbb{R}^n \to \mathbb{R}^{n \times n}\ $ is a real symmetric matrix-valued function of $x$? I.e., $$ x^\intercal A(x) x = 0\ \ \ \forall x \ \ \overset{?}{\implies} \ \ A \equiv 0\ \ \ \forall x $$

Also, is there a name for equations of the form $f(x) = x^\intercal A(x) x$ with symmetric $A$? It is obviously not quadratic since $A$ depends on $x$, but it still has a certain quadratic-form feel.

Answer 1

Nevermind, found a counter-example. The matrix $A$ just needs to have an eigenvector of $x$ with eigenvalue 0, or put another way, $x$ just has to be in the null-space of $A$. This is obvious to construct in 2D, but of course applies generally. Let, $$x := \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

Find a vector orthogonal to $x$ and call it $v$. $$v(x) := \begin{bmatrix} x_2 \\ -x_1 \end{bmatrix}$$

Construct the following matrix by outer-product, $$A(x) := vv^T = \begin{bmatrix} x_2^2 & -x_1x_2 \\ -x_1x_2 & x_1^2 \end{bmatrix}$$

This matrix is symmetric $\forall x$, is not the zero matrix, but still satisfies $x^TAx = 0,\ \ \forall x$.

Answer 2

Starting with what you already know

Imagine a vector $y = Ax$. If $x^T y = 0$, then either $x \perp y$, or either $y$ or $x$ must be $0$. So $A$ must be the composition of a rotation by $\pi / 2$ about some axis in any number of dimensions and any scaling. Just imagine the composition of an arbitrary number of Givens Rotations. We can impose more stringent requirements, but we won't bother.

A rotation other than the identity cannot be symmetric. A fine property of matrices representing rotations is that $A^T = A^{-1}$. If $A = A^T$ then $A=I$ which is not a rotation by $\pi / 2$ about anything.

Fine

So we've shown that $A$ cannot spawn $y \perp x$. Therefore, if $x^T y=0$ then $y = 0$ or $x = 0$. If $y=0$ for any $x \neq 0$ then $A = 0$.

Let's ignore for a moment that $A$ is a function of $x$. If we have any $A$, produced by a drawing from a hat,

Assume there is a function $ A(x) $ for which $ x^T(t) A(x,t) x(t) = 0 $ but $ A(x, t) \neq 0 $. By the above, we are claiming that either:

1. $A(x, t)x(t) \perp x \;\; \forall_{x, t}$

or

2. $A(x, t) = 0$.

We know that 1. must be false, because $A^T = A$, so 2. must be true.