Bilinear form/Finding a basis

Question

Define $B: \mathbb{R^3} × \mathbb{R^3} \to \mathbb{R}$ with $B((x_1,x_2,x_3),(y_1,y_2,y_3)) := -2x_1y_1-x_2y_3-x_3y_2$.

How to check if vectors $v \in \mathbb{R^3}, v \neq 0$ exist such that $B(v,v) = 0$?

I don't know how to start here.

Also, how to find a basis of $\mathbb{R^3}$ such that the Gram matrix of $B$ is in diagonal form with respect to the new basis?

Since $B=\begin{pmatrix} -2 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}$, I got the change of basis $\begin{pmatrix} -\frac{1}{2} & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}$.

But what exactly is the basis in this case?

Answer 1

By definition, for $v=(v_1,v_2,v_3)$, we have $$B(v,v) = -2v_1v_1 - v_2v_3 - v_3v_2.$$ So we get $B(v,v)=0$ for $v=(0,1,0)\neq 0$.

The new basis can be selected by changing variables. Notice that $$B(v,v) = -2v_1^2 - 2v_2v_3.$$ If we make a change of variables $v_1 = x_1$, $v_2 = x_2+x_3$, $v_3 =x_2-x_3$, then the above quantity becomes $$ - 2x_1^2 - 2x_2^2 + 2x_3^2,$$ already in diagonal form. Notice that $(v_1,v_2,v_3)$ is the coordinate of the vector $v$ in the original basis, $(x_1,x_2,x_3)$ is the coordinate of $v$ in the new basis, so what is the new basis? Indeed, $$(v_1, v_2, v_3) = x_1 (1, 0, 0) + x_2(0, 1, 1) + x_3(0, 1, -1),$$ So the new basis consists of the three vectors $$(1,0,0), (0,1,1), (0,1,-1).$$

Answer 2

How to check if vectors $v \in \mathbb{R^3}, v \neq 0$ exist such that $B(v,v) = 0$?

Compute $B(v,v)$ for some $v=(v_1,v_2,v_3)\in\mathbb R^3\setminus\{0\}$. You will see that either $B(v,v)\neq 0$ or you can find a $v$ such that $B(v,v)=0$ holds. Try it!

Also, how to find a basis of $\mathbb R^3$ such that the Gram matrix of $B$ is in diagonal form with respect to the new basis?

Take some base and orthogonalize it with respect to $B$. You can use the Gram Schmidt process. But I don't think it works if $B$ is degenerate.