I ran into the following statement in my lectures but I cannot prove it rigorously.
Statement: Two complex symmetric bilinear forms are equivalent if and only if their ranks are equal.
Definition: We say that two symmetric bilinear forms are equivalent if one of them is obtained from the other by linear change
of variables.
Before this statement we covered the following nice theorem:
Theorem: For any complex symmetric bilinear functional $\xi$ there is a basis such that the matrix of $\xi$ is diagonal with $1$'s and
$0$'s.
So let's try to prove this statement:
$\Rightarrow$ Let $\xi,\eta:V\times V\to \mathbb{C}$ be two bilinear functionals. We can them as the bilinear form:
$\xi(x,y)=X^TAY$ and $\eta(u,v)=U^TBV$, where $X,Y,U,V$ are vector columns and $A=[\xi(e_i,e_j)]$ and $B=[\eta(e_i,e_j)]$ are matrices of $\xi$ and $\eta$, respectively and $\{e_1,\dots,e_n\}$ is some basis of $V$.
Since $\xi$ and $\eta$ are equivalent so suppose that WLOG $\eta$ is obtained from $\xi$ by linear change of variables, i.e.
$$\begin{bmatrix}
x_1 \\
\vdots \\
y_n
\end{bmatrix}=\begin{bmatrix}
d_{11} & \cdots & {d_{1,2n}} \\
\vdots & \ddots & \vdots \\
d_{2n,1} & \cdots & d_{2n,2n}
\end{bmatrix}
\begin{bmatrix}
u_1 \\
\vdots \\
v_n
\end{bmatrix}$$ or shotly $\dbinom{X}{Y}=D\dbinom{U}{V}$ where $D$ is a $2n\times 2n$ matrix with $\det D\neq 0$.
And I don't know what to do next.
Can anyone please give the rigorous proof of this statement (both sides) since I have tried it from yesterday and no results. Would be very grateful for help!
Best Answer
So, what you need to do is to simply use the theorem.
By the theorem, the matrix $A$ is equivalent to a diagonal matrix $D_A$, having only $1$'s and $0$'s, and, by some basic permutation on the elements of the basis, we may suppose that the $1$'s come first in $D_A$. So we have that $$ A = P_A^T \left[\begin{array}{ccccc} I_k & 0 \\ 0 & 0 \end{array}\right] P_A, $$ where $I_k$ is the identity matrix of size $k=rank(A)=rank(\xi)$, for some matrix $P_A$.
Analogously, we have that $$ B= P_B^{T}\left[\begin{array}{ccccc} I_m & 0 \\ 0 & 0 \end{array}\right] P_B, $$ with $m=rank(B)=rank(\eta)$.
If $rank(\xi)=rank(\eta)$, then $k=m$, and consequently, $$ A= P_A^{T}\left[\begin{array}{ccccc} I_m & 0 \\ 0 & 0 \end{array}\right]P_A = P_A^{T}(P_B B P_B^{T})P_A = (P_B^{T}P_A)^T B P_B^{T}P_A. $$
Conversely, suppose that $\xi$ and $\eta$ are equivalent, that is, there exists some matrix $N$ such that \begin{align} A= N^T B N & \Rightarrow \ P_A^{T}\left[\begin{array}{ccccc} I_k & 0 \\ 0 & 0 \end{array}\right]P_A = N^T P_B^{T}\left[\begin{array}{ccccc} I_m & 0 \\ 0 & 0 \end{array}\right] P_B N \\ &\Rightarrow \left[\begin{array}{ccccc} I_k & 0 \\ 0 & 0 \end{array}\right] = P_AN^T P_B^{T} \left[\begin{array}{ccccc} I_m & 0 \\ 0 & 0 \end{array}\right] P_B N P_A^T \\ & \Rightarrow \left[\begin{array}{ccccc} I_k & 0 \\ 0 & 0 \end{array}\right]= (P_B N P_A^T)^T \left[\begin{array}{ccccc} I_m & 0 \\ 0 & 0 \end{array}\right] P_B N P_A^T. \end{align} But then it follows from the uniqueness of the rank of a bilinear form that $rank\left(\left[\begin{array}{ccccc} I_k & 0 \\ 0 & 0 \end{array}\right]\right)=rank\left(\left[\begin{array}{ccccc} I_m & 0 \\ 0 & 0 \end{array}\right]\right)$, and consequently, $k=m$, that is, $rank(\xi)=rank(\eta)$.