Bilinear forms diagonalization

Let $V$ be a vector space, and let $ϕ$ be a bilinear form

ϕ : V \times V ⟶ R

If $ϕ$ is symmetric, then given a basis $B$ we get a matrix $M_{B}$ that is symmetric. For $v, w \in V$ , and following the characterization here we get:

ϕ (v, w) = b^{- 1} (v)^{t} \cdot M_{B} \cdot b^{- 1} (w)

Since $M_{B}$ is symmetric, we can apply successive transformations like

(\begin{array}{cccc} 1 & 0 & \dots & 0 \\ - \frac{a_{12}}{a_{11}} & 1 & ⋮ \\ ⋮ & ⋱ & 0 \\ - \frac{a_{1 n}}{a_{11}} & 0 & \dots & 1 \end{array}) \cdot (\begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{12} \\ ⋮ & M \\ a_{1 n} \end{array}) \cdot (\begin{array}{cccc} 1 & - \frac{a_{12}}{a_{11}} & \dots & - \frac{a_{1 n}}{a_{11}} \\ 0 & 1 & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & 0 & 1 \end{array}) = (\begin{array}{ccccc} a_{11} & 0 & \dots & 0 \\ 0 \\ ⋮ & M^{'} \\ 0 \end{array})

And so we get a diagonal matrix $D$ that verify:

D = P^{t} M_{B} P

If we considerate a new basis $N$ such that $n = b \circ P$ , it is clear that the coordinates of $v$ and $w$ are $n^{- 1} (v) = P^{- 1} b^{- 1} (v)$ and $n^{- 1} (w) = P^{- 1} b^{- 1} (w)$ . Since

ϕ (v, w) = b^{- 1} (v)^{t} \cdot M_{B} \cdot b^{- 1} (w) = n^{- 1} (v)^{t} \cdot D \cdot n^{- 1} (w)

we conclude that $D$ is the matrix of $ϕ$ in the new basis $N$ .

From here, with an easy change of basis, rearranging the elements and scaling them by $\frac{1}{\sqrt{a_{i i}}}$ , we obtain that the matrix is of the form:

{(D)}_{i j} = {\begin{array}{cl} 1 & si 1 \leq i \leq r, j = i \\ - 1 & si r + 1 \leq i \leq s, j = i \\ 0 & otherwise \end{array}

The integers $r$ and $s$ are constant through base changes and they allow us to define the signature of the bilinear forms.