# Thread: Find a degenerate and a non-degenerate bilinear form

1. ## Find a degenerate and a non-degenerate bilinear form

Hello,

My question is,

A bilinear form $w$ on $U \oplus V$ is degenerate if, as a function of one of its two arguments, it vanishes identically for some non-zero value of its other argument; otherwise it is non-degenerate.

(a) Give an example of a degenerate bilinear form (not identically zero) on $\mathbb{C}^2 \oplus \mathbb{C}^2$.

(b) Give an example of a non-degenerate bilinear form on $\mathbb{C}^2 \oplus \mathbb{C}^2$.

The definition we have for a bilinear form is: $w$ on $W=U \oplus V$ is a bilinear form if, $$w(a_1x_1 +a_2 x_2, y)=a_1 w(x_1, y)+a_2w(x_2,y) \: \text{and} \: w(x, a_1 y_1+a_2 y_2)=a_1w(x,y_1)+a_2w(x, y_2).$$ And the only example we have of a bilinear form is the dot product in $\mathbb{R}^n \oplus \mathbb{R}^n \to \mathbb{R}$.

I'm not entirely sure what is so confusing here for me. I want to just take my only example, adapt it for a 2 dimensional case, and shove in complex entries for each array. So, $$\begin{pmatrix}x_1\\x_2\end{pmatrix}^T \begin{pmatrix}y_1 \\ y_2 \end{pmatrix}=x_1y_1+x_2y_2,$$ now just set this as being equal to zero. But that doesn't make any sense to me.

The only other thing I have is, $$w(x_1 + x_2 , y_1 + y_2)=w(x_1, y_1)+w(x_1, y_2)+w(x_1,y_1)+w(x_2,y_2),$$ which is just the full definition expanded out. But this seems to be too much as the question wants to restrict attention to "as a function of one of its two arguments".

So I'm not sure what to do.

2. ## Re: Find a degenerate and a non-degenerate bilinear form

Hey bkbowser.

Have you got the axioms of a bi-linear form?

You will have to show that given those axioms you can pick a vector such that it vanishes or show that it doesn't.

3. ## Re: Find a degenerate and a non-degenerate bilinear form

I have only, $$w(a_1x_1+a_2x_2,y)=a_1w(x_1,y)+a_2w(x_2,y),$$ and a definition of a linear functional, which would be $y$ in the previous, as $y(x)=b_1 y_1 + \dots + b_n y_n$, for some set of scalars $\{b_i\}$.

Since $x, y \in \mathbb{C}^2$ I should be able to write this out as, $$w(x,y)=w((a+bi,c+di),(e+fi,g+hi))=(a+bi)(e+fi)+( c+di)(g+hi),$$ and I just want to know when this is equal to zero. So expand it, $$ae+afi+bei + bfi^2+cg+chi+dgi+dhi^2.$$ Or I guess you can look at when, $$(a+bi)(e+fi) =-(c+di)(g+hi),$$ for some fixed pair $e+fi$, $g+hi$ with $a+bi$, and $c+di$ allowed to vary over all of $\mathbb{C}$.

4. ## Re: Find a degenerate and a non-degenerate bilinear form

Try fixing one vector to be the zero vector and show that if the other is varied then you will either get a non-zero answer or a zero answer.

If you get a zero answer then it is non-degenerate and if not it is degenerate.

5. ## Re: Find a degenerate and a non-degenerate bilinear form

Define $w(x,y)=x_1y_1$ where $x=(x_1,x_2)\in \mathbb{C}^2$, and $y=(y_1,y_2)\in \mathbb{C}^2$. Since, $$w(c\alpha+\beta),y = (c\alpha_1+\beta_1)y_1=c\alpha_1 y_1 +\beta_1 y_1 = w(c\alpha,y)+w(\beta,y),$$ it's a bilinear form.

If you set $y=(0,1)$ then $y \neq 0$ but $w(x,y)=x_1\cdot 0 =0$ for all $x \in \mathbb{C}^2$. So this satisfies (a).

If instead $y=(1,0)$ then $y \neq 0$ and $w(x,y)=x_1\cdot 1 =0 \iff x_1=0$? I don't think that quite sats (b) but I should be able to make something work.