Let $V,W$ be vector spaces over field $\mathbb F$.
Let there be two functionals; $\psi : V \to \mathbb F$ , $\phi : W \to \mathbb F$.
How to find the Canonical form (of orthogonal matrix, meaning it has either $+1,-1$ on the diagonal or $2\times 2$ squares of rotations) of the next bilinear form: $$f: W\times V \to \mathbb F$$ that is defined: $$\forall v,w \in V,W \ \ f(w,v) = \phi(w)\psi(v)$$
I know I need to build specific bases for $\psi , \phi$ for which there will be a canonical form. Also, lets assume that both $\phi ,\psi \ne 0$, otherwise there is nothing special to do and the canonical form is just zeros.
Would love some ideas on that, thanks!
Note that the kernels (nullspaces) of $\psi,\phi$ have dimensions $\dim(V) - 1$ and $\dim(W) - 1$ respectively. Let $v_2,\dots,v_{n}$ and $w_2,\dots,w_{m}$ be bases of these respective kernels. Extend these bases so that $\{v_1,\dots,v_n\}$ is a basis for $V$ and $\{w_1,\dots,w_m\}$ is a basis of $W$; scale $v_1,w_1$ so that $\psi(v_1) = \psi(w_1) = 1$. These bases are such that $$ f\left(\sum a_i v_i,\sum b_jw_j \right) = \pmatrix{a_1\\ \vdots \\ a_n}^T \pmatrix{1&0&\cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots &\vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0}\pmatrix{b_1 \\ \vdots \\ b_m} = a_1b_1 $$