Prove that bilinear forms are the same if and only if $A=B$

Question

I am looking for help or a direction to prove that $x^TAy=x^TBy$ if and only if $A=B$.

So far I've been using inner product and trace, for example for quadratic form I've used trace to prove $x^TAx=x^TBx$ if and only if $A+A^T=B+B^T$, but how is bilinear form different, shouldn't same approach apply here.

Lets say: $$x^TAy=x^TBy$$ $$tr(x^TAy)=tr(x^TBy)$$ $$tr(x^TAy)-tr(x^TBy)=0$$ $$tr(x^T(A-B)y)=0=tr(x^T(A-B)^Ty)$$ which means that the solution with either be $ A=B $ or $(A-B)$ is skew symmetric $(trace=0)$

What am I doing wrong here? It should be only if $A=B$

Answer 1

You want to have $x^T (A - B) y = 0$ for all $x, y \in \mathbb{R}^n $

Using the standard basis, we can write:

${e_i}^T (A - B) {e_j} = 0 $ for all $i, j$,

but $(e_i^T (A - B) e_j )$ is just the $i,j$-th entry of $(A - B)$

Hence $A - B$ must be the zero matrix, from which it follows that $ A = B$