Let $V$ be a finite dimensional vector space over $\mathbb C$. Suppose $f,g$ are symmetric bilinear forms on $V$. Prove there's an invertible $T:V\to V$ such that $f(Tx,Ty)=g(x,y)$ iff $f,g$ have equal rank. Is the claim true over $\mathbb R$ as well?
I thought the answer over both $\mathbb R$ and $\mathbb C$ is yes, by taking $T:b_i\to c_i$ for $(b_i),(c_i)$ respectively orthonormal bases for $f,g$. However, I don't even know how to prove my claim, and suspect I'm missing something important over $\mathbb R$...
For the first part: you can always diagonalize a complex bilinear form to have just ones and zeros down the diagonal. The rank will be equal to the number of $1$s. If two forms diagonalize to the same matrix, there is a change of basis between the two.
For real forms, you can diagonalize down to $0$ and $\pm 1$. Now two forms whose Gram matrices have equal rank can still have different signatures. For example, the bilinear form $f(x,y)=xy$ can't be changed to the bilinear form $g(x,y)=-xy$.