Can anybody help me in construction of linear operators in exercise 12 of section 10.1, Hoffman and Kunze.
Let $f,g$ be bilinear forms on a finite dimensional vector space $V$. Suppose $g$ is non singular. Show that there exist unique linear operators $S,T$ on $V$ such that $f(a,b)=g(Sa,b)=g(a,Tb)$ for all $a,b$. What if $g$ is singular?
I am not getting how to start. Any hint will be appreciated. Thanks.
Let $(e_k)$ be a basis for $V$, and let $A$ and $B$ be the matrices representing $f$ and $g$ relative to this basis; thus (for example) $$ f(a,b) =\sum_{i,j} a_i A_{ij}b_j, $$ where $a=\sum_i a_ie_i$ and $b=\sum_{j}b_je_j$. Because $g$ is nonsingular, the matrix $B$ is nonsingular. Let $T$ be the linear operator whose matrix relative to the basis $(e_k)$ is $B^{-1}A$. I claim that $f(a,b)=g(a,Tb)$ for all $a,b\in V$. To show this is a pretty routine calculation. The uniqueness of $T$ will follow from the nonsingularity of $g$. The construction of $S$ is similar.