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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)?

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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 $\mathbb{R}^2$.

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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.

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