Bilinear Form and Rank Proof I found a task in the internet about bilinear forms that I would like to unterstand. I changed a few variables but it goes like this. Suppose that $\psi: U \times V \rightarrow K$ is a bilinear form of rank $r$ on finite dimensional vector spaces $U$ and $V$
over $K$. I want to show that there exist bases $e_1, . . . , e_m$ for $U$ and $f_1, . . . , f_n$ for $V$ such that
$$\psi \big{(}\sum_{i=1}^m x_ie_i, \sum_{j=1}^n y_jf_j \big{)}=\sum_{k=1}^r x_ky_k$$ for all $x_1, . . . , x_m, y_1, . . . , y_n \in K$. Then how to find the dimensions of the left and right kernels of $\psi$? Thanks in advance for advice of any kind!
 A: Let
$$\begin{array}{l|rcl}
\psi_1 : & U & \longrightarrow & V^* \\
    & u & \longmapsto & (v \mapsto \psi(u,v)) \end{array}$$
The rank of $\psi_1$ is equal to the one of $\psi$, i.e. equal to $r$. Let $\{f_1^*, \dots, f_r^*\}$ be a basis of $\text{im}\ \psi_1$ that we complement into a basis
$\{f_1^*, \dots, f_r^*, f_{r+1}^*, \dots, f_n^*\}$ of $V^*$. We can find $e_1, \dots , e_r \in U$ such that $\psi_1(e_i) = f_i^*$ for $1 \le i \le r$. $\{e_1, \dots , e_r\}$ is linearly independent as $\{f_1^*, \dots, f_r^*\}$ is. We again complement $\{e_1, \dots , e_r\}$ into a basis $\{e_1, \dots , e_r, e_{r+1}, \dots, e_m\}$ of $U$. $\{e_{r+1}, \dots, e_m\}$ is a basis of $\ker \psi_1$.
Now let $\{f_1, \dots, f_n\} \subseteq V$ be the dual basis of $\{f_1^*, \dots, f_r^*, f_{r+1}^*, \dots, f_n^*\}$.
For $1 \le i \le r$  and $j \in \{1,\dots, n\}$ we have
$$\psi(e_i,f_j) = (\psi_1(e_i))(f_j)=f_i^*(f_j)=\delta_{ij}$$ where $\delta_{ij}$ stands for Kronecker delta.
While for $i \gt r$  and $j \in \{1,\dots, n\}$
$$\psi(e_i,f_j) = (\psi_1(e_i))(f_j)=0(f_j)=0.$$
Which allows to conclude to the desired result by bilinearity of $\psi$.
Note: I implicitly proved the result mentioned in jlammy response that is called the rank normal form of a matrix.
A: Let $M_{ij}=\psi(e_i,f_j)$ be the matrix representing the bilinear form, in the bases $\{e_i\}$ and $\{f_i\}$. It is a theorem that we can choose bases so that the form of $M$ is
$$M=\begin{pmatrix}I_r&0\\0&0\end{pmatrix}.$$
Then in these bases it is clear that $\psi(x,y)=x^\top My$ acts as you want.
The left kernel is then the span of $e_{r+1},\dots,e_m$ (corresponding to those bottom rows of zeros in the matrix), so the dimension is $m-r$. Similarly the right kernel is the span of $f_{r+1},\dots,f_n$ which has dimension $n-r$.
