I am introducing myself to bilinear forms. I would like a little help to do this exercise:
Let $f$ and $g$ two bilinear forms on $V$, with $f$ non-degenerated. Show that there is an unique operator $T\in End(V)$ such that $g(x,y)=f(x,T(y))$ for all $x,y \in V$. Show that T is biyective if and only if g is also non-degenerate.
I think I can choose a basis of V to express the matrix of $T$ in terms of the matrices of $f$ and $g$, but I am not completely sure. Thank you!
It might help to think about what this is saying in the case $V=\Bbb R^n$. In this case there are matrices $A_f$ and $A_g$ such that \begin{align*} f(x,y) &= x^\top A_f y & g(x,y) &= x^\top A_g y \end{align*} Moreover, $A_f$ is invertible since $f$ is nondegenerate.
Now, an endomorphism of $\Bbb R^n$ is simply an $n\times n$ matrix $B$. Then $$ f(x,By)=x^\top A_f(By)=x^\top (A_fB)y $$ Note that chosing $B=A_f^{-1} A_g$ gives $$ f(x,By)=x^\top A_g y= g(x,y) $$ Do you see why this choice is unique? Do you see how to extend this argument to the general case?