A quadratic form over $K-$vector space $V$

Question

Let $K$ a field, $\operatorname{char} K \ne 2$.

Definition: A quadratic form over $K$ is a homogeneous polynomial $Q(x_1, x_2, \dots , x_n) \in K[x_1, x_2, \dots , x_n]$ of degree $2$.

If $V\cong K^n$, then each quadratic form defines a function $Q:V \rightarrow K$.

In General:

Definition: A quadratic form over $K-$vector space $V$ ($dim_K V=n$) is a function $Q:V \rightarrow K$ where, if $V$ has as basis the $\{e_1, e_2, \dots , e_n\}$ then $Q: K^n \rightarrow K$ is defined by $Q(x_1 e_1+ \dots x_n e_n)$ and is given from a quadratic form.

---

What do the following mean?

• A quadratic form over $K-$vector space $V$ ?

• Can you explain to me the second definition ?

A $K-$bilinear form over $K-$vector space $V$ is a function that is $B: V \times V \rightarrow K$ linear in respect to each variable.

$K-$bilinear form $B: V \times V \rightarrow K$ is called symmetric $\Leftrightarrow B(v, w)=B(w, v), \forall v, w \in V$.

There is a bijective mapping between: 1. the quadratic forms over $V$ 2. the bilinear forms over $V$

$$Q(x) \rightarrow B(x, y)=\frac{Q(x+y)-Q(x)-Q(y)}{2}$$

$$B(x, y) \rightarrow Q(x)=B(x, x)$$

Can you explain to me the mapping

$$Q(x) \rightarrow B(x, y)=\frac{Q(x+y)-Q(x)-Q(y)}{2}$$

It can be proven that there is an unique quadratic matrix $A \in M_n(K)$ such that $Q(x)=x^T A x$ and $B(x, y)=x^T A y$.

The order of the quadratic form is defined to be the order of the matrix $A$.

Can you explain to me how we can prove that there is an unique quadratic matrix $A \in M_n(K)$ such that $Q(x)=x^T A x$ and $B(x, y)=x^T A y$?

Edit:

$$(i) B(v_1+v_2, w)=B(v_1, w)+B(v_2, w)$$ $$(ii) B(\lambda v, w)=\lambda B(v, w)$$

$$B(\vec{x}, \vec{y})=B \left ( \sum_{i=1}^n x_i \vec{e}_i, \sum_{j=1}^n x_j \vec{e}_j \right )\overset{ (i) }{= } \sum_{i=1}^nB \left ( x_i \vec{e}_i, \sum_{j=1}^n x_j \vec{e}_j \right )\overset{ (i) }{= } \sum_{i=1}^n \sum_{j=1}^n B \left ( x_i \vec{e}_i, x_j \vec{e}_j \right ) \\ \overset{ (ii) }{= } \sum_{i=1}^n \sum_{j=1}^nx_i B \left ( \vec{e}_i, x_j \vec{e}_j \right )\overset{ (ii) }{= } \sum_{i=1}^n \sum_{j=1}^nx_i x_j B \left ( \vec{e}_i, \vec{e}_j \right )= \sum_{i=1}^n \sum_{j=1}^nx_i x_j a_{ij}$$ where $$a_{ij}=B \left ( \vec{e}_i, \vec{e}_j \right )$$ Since $B$ is symmetric we have that $$a_{ij}=B \left ( \vec{e}_i, \vec{e}_j \right )=B \left ( \vec{e}_j, \vec{e}_i \right )=a_{ij}$$ so we have that $A=(a_{ij})$ is symmetric. Why can we write $$B(\vec{x}, \vec{y})=\vec{x}^T A \vec{y}$$ ???

Answer 1

OK, there's a lot here, so let's try and tackle one thing at a time (in general it's best to break up big questions like this into a couple of smaller ones).

• A quadratic form over a vector space means that $q$ is a function which takes vectors as arguments, it just means it has more than one variable, that's all.

• What exactly don't you get? The definition is that it's a function, that is a rule which does the following:

$$\mathbf{v}=\sum_{i=1}^n x_i \mathbf{e}_i\mapsto q(x_1,x_2,\ldots, x_n)$$

i.e. it takes the linear coefficients of the basis vectors, $\mathbf{e}_i$ and returns the value of the quadratic form where you plug in the coefficients for the variables.

• Again, you need to say what you mean by "explain," that's exceedingly vague. The idea is that you can turn any quadratic form into a bilinear form using that formula. Take, for example, $q(x)=2x^2$, over $V=K$, the one-dimensional vector space. Then the bilinear form associated to this is

$$B(x,y)={q(x+y)-q(x)-q(y)\over 2}={3(x+y)^2-3x^2-3y^2\over 2}= 3xy$$

• The quadratic matrix is easy, any quadratic form has the form

$$\sum_{i=1}^n\sum_{j=1}^n a_{ij}x_ix_j$$

then the matrix for $q$ is just $A=[a_{ij}]$ and the matrix for $B$ is the same.