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1. Bilinear forms are functions $$Q:U\times U\to k$$ that depend on each of the arguments linearly.

Alternatively, one can think of them as the linear operators
$A:U\to U^*, \mathrm{ \ with\ } Q(u,v)=A(u) (v).$
If $$U$$ has a basis, the bilinear form can be identified with the matrix of its coefficients:
$Q_{ij}=Q(e_i,e_j).$

(Notice that the order matters!)

Rank 1 bilinear forms of rank 1 are just products of linear functions, $$\bra{u}\bra{v}$$.

2. In the space of continuous functions on interval $$[a,b]$$, a kernel $$K: [a,b]\times[a,b]\to\Real$$ defines a bilinear form
$Q(f,g)=\int_a^b \int_a^b K(s,t)f(s)g(t) ds dt.$

3. Bilinear form can be symmetric, $$Q(u,v)=Q(v,u)$$, or skew-symmetric $$Q(u,v)=-Q(v,u)$$, and each bilinear form is a sum of symmetric and skew-symmetric forms. (Skew-)symmetric form correspond to (skew-)symmetric matrices.

4. Passing to a new basis is straightforward:
$Q\mapsto C^\top Q C,$
where $$C$$ is the matrix of passing from new to old basis, abd $$C^\top$$ is its transpose.

5. Given a bilinear form $$Q$$, one can derive a quadratic form $$q(x):=Q(x,x)$$. Adding a skew-symmetric form to a bilinear form leave the associated quadratic form unchanged. Hence, one can assume that a quadratic form comes from a symmetric bilinear form, which can be derived using the polarization process:
$Q(x,y)=\frac12(q(x+y)-q(x)-q(y)).$

6. A quadratic form on a real vector space is called positive definite is
$q(x)>0 \mathrm{\ for\ any\ } x\neq 0.$

An example of positive definite quadratic form is the standard Euclidean norm (in a basis $$\{e_k\}_{k=1}^n$$):
$q(x)=\sum x_k^2.$
One can use any positive definite quadratic form as the defining building block for a Euclidean space (define the norm of a vector as $$\|x\|^2:=q(x)$$, the scalar product as $$(x,y):=Q(x,y)$$ etc). Any theorem about the Euclidean spaces can be reformulated and proven in terms of a positive-definite quadratic form without any loss (say, the Cauchy-Schwartz inequality
$(x,y)\leq \|x\|\|y\|$
translates into
$Q(x,y)\leq \sqrt{q(x)q(y)}.$

Still, one needs convenient coordinates. Normal forms in the case of positive-definite forms is done by the familiar process (Gram-Schmidt orthogonalization). The process is straightforward: given a basis $$f_k, k=1,\ldots, n$$, set $$e_1:=f_1$$, so that and then iterate for $$k\gt 1$$:
$e_k=f_k+\sum_{l\lt k}c_{kl}e_l,$
where the coefficients $$c_{kl}$$ are chosen so that $$Q(e_k,e_l)=0$$ for $$k<l$$, i.e.
$c_{kl}=-\frac{Q(f_k,e_l)}{Q(e_l,e_l)}.$
(Here we use the fact that $$e_l\neq 0$$ – thanks to the linear independence of $$f$$’s, – and positive definiteness of $$Q$$.)

Exercise: Consider the space of real polynomial functions of degree at most 3. Consider the form
$q_1(f):=|f(-1)|^2+|f(0)|^2+|f(1)|^2.$
Is this form positive definite?

Consider another form,
$q(f)=\int_0^\infty e^{-x}|f(x)|^2dx.$
Diagonalize the form using Gram-Schmidt procedure starting with the standard monomial basis $$\{1,x,x^2,x^3\}$$.

7. Of course, not only positive definite quadratic forms can be brought to simple normal form; Jacobi diagonalization process works always. It creates a sequence of linear functions $$e^*_k, k=1,\ldots,n$$ (that is elements of the dual space) such that
$q(v)=\sum a_l e_k(v)^2.$

The procedure works like this: in the original basis, the form is
$\sum_{k,l} Q_{kl}x_kx_l.$
If $$Q_{11}\neq 0$$, one can represent
$q(x)=Q_{11}(x_1+\sum_{l\geq 2}Q_{1l}/Q_{11} x_l)^2+q^{(1)}$,
where $$q^{(1)}$$ depends only on coordinates $$x_2,\ldots,x_n$$. Set $$y_1=x_1+\sum_{l\geq 2}Q_{1l}/Q_{11} x_l$$; continuing inductively, we get a diagonalized quadratic form.

If at any step all diagonal elements $$Q_{ll}=0$$, change coordinates to $$x_l-x_m, x_l+x_m$$.

8. The numbers $$n_+, n_-$$ of positive and negative coefficients in a diagonal representation of the quadratic form are called its signature. The signature is an invariant of a quadratic form (that is it does not matter, what is the diagonalization process, the signature of the result is always the same. In fact, $$n_+$$ ($$n_-$$) is the largest dimension of a subspace restriction to which of the quadratic form is positive (negative) definite. At the same time, $$n_0:=n-n_+-n_-$$ is the dimension of $$\ker Q$$.

9. The diagonalization process can be made perfectly efficient is the principal minors of the Gram matrix, that is the determinants
$\Delta_0:=1, \Delta_1:=Q_{11}, \Delta_2=\left|\begin{array}{cc}Q_{11}& Q_{12}\\Q_{21}&Q_{22}\\\end{array}\right|,\ldots$
are non-vanishing. In this case, the diagonalization results in the quadratic form
$\sum_{k=1}^n\frac{\Delta_k}{\Delta_{k-1}}y_k^2,$
implying the Sylvester criterion for a quadratic form to be positive definite: all principal minors of its Gram matrix (with respect to any basis) should be positive.

Exercise. Find the signature of the quadratic form $$q_1$$ above.