TITLE: Non-degenerate bilinear forms
QUESTION [6 upvotes]: Let $b$ be a non-degenerate bilinear form on a finite dimensional vector space $V$. Let $b'$ be any bilinear form on $V$. Show that $\exists$ $T \in L(V,V)$ such that $b'(v,w)=b(Tv,w) \,  \forall v,w \in V$.

REPLY [3 votes]: Let $B$ and $B'$ be matrices such that $xBy^T=b(x,y)$ and $xB'y^T=b'(x,y)$.
Since $b$ is nondegenerate, the rows of $B$ are linearly independent, so there exists a matrix $T$ such that $TB=B'$. Can you take it from here?