# Question on bilineal and quadratic forms

My first question is:

Let $$\phi: V \rightarrow \mathbb{R}$$ a quadratic form. We say that a vector $$x$$ is autoconjugate if its conjugate with its self, that is, $$\phi(x)=0$$. Does the set of all autoconjugate vectors form a subspace of V? (My guess: The set of all those vectors would be :$$S= \{x \in V | \phi(x) = 0\}$$ which is extremely similar to the Ker of something. That is where my problem lies, does a quadratic form have such a thing like a Ker? Is it equal to the Ker of the bilinear form or does a bilinear form have a Ker?

My second question is:

How do bilinear and quadratic forms look like (I mean visually)?

You can define that subset like you just did, but in general it'll only be a cone (that is, $$\lambda x \in S$$ for any scalar $$\lambda$$ and $$x \in S$$).

For example, take the quadratic form $$\phi(x,y) = x^2 - y^2$$ on $$\mathbb{R}^2$$. Then $$S = \{ (x,y) \mid x = y \text{ or } x = -y \}.$$ This is the union of two lines, but is not a linear subspace of $$\mathbb{R}^2$$.

I don't have the full answer, but I believe you can define a Ker for quadratic and bilinear forms, yes. They just might not have the properties they would have for linear operators, such as being a vector space for instance.

So even if the answer to your problem is indeed a Ker, I'm not sure it helps a lot to know it.

As for what they look like, well the simplest quadratic form is a degree 2 polynomial, and one of the most common bilinear forms is a scalar product, even if it's a bit harder to visualise.