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Let $v \in T_2(V)$ be a bilinear form over finite space V. Let T be a Linear transformation $V \to V$. We define: $v_T(x,y) = v(T(x),y)$.

Assuming $v$ is nondegenerate, let us have another bilinear form $\xi \in T_2(V)$. Prove that there exists exactly one transformation $T$ so $\xi = v_T$.

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My idea is trying to find T explicitly, by showing that if $v_T$ and $\xi$ operate the same on each two vectors in the basis $(e_i,e_j)$, than they must be equal.

Taking $v_T(e_i,e_j) = \sum_{i,j}\alpha_{ij}T(e_i)e_j = \sum_{i,j}\beta_{ij}e_ie_j = \xi(e_i,e_j)$ and defining $T$ so $T(e_i) = \begin{pmatrix} \frac{\beta_{i1}}{\alpha_{i1}}\\ \frac{\beta_{i2}}{\alpha_{i2}}\\ \vdots \\ \frac{\beta_{in}}{\alpha_{in}}\\ \end{pmatrix}$ Thus making both of the sides equal.

I'm not sure whether this approach is correct. Also, I believe I used the nondegenerate fact at the point that for every vector in the basis $v_T \neq 0$ but I'm not sure exactly how it works formally.

Would love your help getting directions to solve this, and comments about my approach, thanks!

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By using the fact that $(AB)^T = B^TA^T$ (where $^T$ denotes a transposed matrix), we get: $$[x]^T[v_t][y] = ([T][x])^T[v][y] = [x]^T[T]^T[v][y] =[x]^T([v]^T[T])^T[y] $$ So, that brings us to:$$ [\xi]^T = [v]^T[T]$$

Notice $[v]$ is invertible, as $v$ is non-degenerate, so: $$ [T]=([v]^T)^{-1}[\xi]^T$$

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