Be $f:V \times V \to F$ a bilinear pattern and $V$ of finite dimension. Is it correct that for every linear transformation $T:V \to V$ exists another linear transformation $T':V \to V$ for which: $$f(Tu,\ v) = f(u,\ T'v),$$ for every $v,u$ of $V$?
I think the way should be defining $B$ an orthonormal base for $V$ and representing $f$ by $B$.
I'm having trouble with all the Sigma additives and would be happy to get some help with this.
Thanks,
You need some assumption on the bilinear form $f$. For a general form the claim is not necessarily true.
However, if we assume that $f$ is nondegenerate, the claim holds. Being nondegenerate means that for every $v\in V$ there is some $u\in V$ with $f(v,u)\ne0$. Alternatively, you can choose a basis for $V$ and represent $f$ with respect to this basis by a matrix $A$, and then being nondegenerate means that $A$ is nonsingular.
If $f$ is represented by $A$, it means that for every $u,v\in V$ we have $$f(u,v)=u^tAv,$$ where we write $u,v$ as coordinate vectors, with respect to the same above chosen basis. If $A$ is nonsingular, then for every $u_1,\ldots,u_n$ which are linearly independent, $u_1^tA,\ldots,u_n^tA$ are also linearly independent. Now say we want to determine $T'(v)$ for some $v\in V$. If we think through coordinates, the desired $T'(v)$ is a column vector, and in order to find it we need $n$ independent linear equations. We take a basis $u_1,\ldots,u_n$ and for every $i=1,\ldots n$ we have a linear equation in $T'(v)$, namely $$u_i^tAT'(v)=f(T(u),v).$$ The nonsingularity assures us that these equations are indeed independent, and thus have a unique solution, thus $T'(v)$ is well defined.