Inner product and canonical forms If $V$ is a vector space with finite dimension and for some symmetric bilinear form $f: V \times V \rightarrow \mathbb{R}$, how to I show that $f$ defines an inner product iff the unique real canonical form of $f$ is $I_n$.
So I know that we can represent the bilinear form as a quadratic form and from sylvesters law of inertia we have that there must be exactly one canonical (Thus proving uniqueness). However, for this how do I start showing that the basis elements are all orthogonal with full rank?
Any help would be greatly appreciated!
 A: Take any basis $\{{e_i\}}$ of $V$; there should be no problem in constructing one. This enables one to construct the symmetric matrix: $f_{i j}:=f(e_i,e_j)$. Diagonalise $f_{i j}$ using the eigenvalues. If all the eigenvalues are positive, it is an inner product.
A: First you construct the matrix $A = (f(e_i,e_j))$ for any given basis $\{e_1, \dots, e_n\}$ of $V$. By Sylvester's law of inertia, there exists an invertible matrix $S$ such that $B=S^TAS$ is diagonal with only $-1,0,1$ as eigenvalues. 
To be a inner product $f$ must satisfy some properties. We already have that $f$ is symmetric and bilinear, it remains to impose that $f$ is positive definite: $f(x,x) \geq 0$ and $f(x,x)=0 \Leftrightarrow x=0$. This happens if and only if $B$ is positive definite which means that the only possible eigenvalue for $B$ is $1$. 
To finish, let $\{v_1, \dots, v_n\}$ be the basis given by $v_j = Se_j$. In this new basis $f$ is the standard inner product.
Another way to see this is to use the spectral theorem on $A$ and then rescale the eigenvalues. You'll actually prove Sylvester's theorem along the way. 
