Consider Bilinear Forms, and two real $n \times n$ matrices $B$ and $\tilde B$.
Suppose we have that
$$(Bx,x) = (\tilde B x, x)$$
for all $x\in \mathbb R^n$.
What can we can say about $B-\tilde B$?
(Note that there is no assumption of symmetry for either matrix.)
My thoughts:
From properties of the inner product $<.,.>$ we know that if
$$<x,y> = <z,y>$$
for all $y\in \mathbb R^n$, then $x=z$.
Can I use this same property and look at the bilinear form equation above and also claim that $Bx = \tilde Bx$?
Then
$$Bx - \tilde Bx = 0$$ $$\implies (B-\tilde B)x = 0$$ $$\implies (B-\tilde B)=0_{nxn} $$
This is apparently not true, and that from the bilinear form equation above, one concludes that $(B-\tilde B)$ is skew-symmetric.
Where have I gone wrong? How can I show the skew-symmetry?
Any ideas are welcome.
Thanks,
You cannot conclude directly to the skew symmetry of the matrix $A=B-\tilde B$ because you have the same variable vector $x$ in the bilinear form $$(Ax,x)=0.$$
In order to conclude, you can proceed as follows. For any two vectors $x,y$, you have $$\begin{aligned}0=(A(x+y),x+y)&=(Ax+Ay,x+y)\\ &=(Ax,x)+(Ax,y)+(Ay,x)+(Ay,y)\\ &=(Ax,y)+(Ay,x)\end{aligned}$$ Which proves that $A$ is skew symmetric providing that $(\cdot,\cdot)$ is nondegenerate.