If $x^TAx$ = $x^TBx$ for all $x\in \mathbb R^n$. Then what can I say about the matrices? Are they congruent to each other?

Question

If $x^TAx$ = $x^TBx$ for all $x\in \mathbb R^n$. Then what can I say about the matrices? Are they congruent to each other?

My attempt: $x^TAx$ = $x^TBx$ for all $x\in \mathbb R^n$. Then I can say $A - B$ is skew-symmetric. I cannot see more than this.

Can anyone please help me?

Answer 1

Some terminology: for a matrix $A$, we say that a function $Q(x)=x^TAx$ is a quadratic form generated by $A$. As you can see, quadratic form depends only on the symmetric part of the matrix.

The statement

$A$ and $B$ generate the same quadratic form, that is $x^TAx=x^TBx$ for every $x\in \mathbb R^n$.

is equivalent to

$A-B$ is skew-symmetric.

In particular, they do not need to be congruent - for example the following matrices generate the same quadratic form ($Q(x)=0$): $$\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix},$$ but surely they are not congruent as their determinants differ.

Answer 2

With $A=0$ and $B$ skew-symmetric you have $\forall x$: $$x^t Ax=0=x^t Bx$$ but $A$ and $B$ are not necessarily congruent.