Dania Mueller

## Answered question

2022-06-27

Let $v\in {T}_{2}\left(V\right)$ be a bilinear form over finite space V. Let T be a Linear transformation $V\to V$. We define: ${v}_{T}\left(x,y\right)=v\left(T\left(x\right),y\right)$
Assuming $v$ is nondegenerate, let us have another bilinear form $\xi \in {T}_{2}\left(V\right)$. Prove that there exists exactly one transformation $T$ so $\xi ={v}_{T}$.

### Answer & Explanation

Tianna Deleon

Beginner2022-06-28Added 29 answers

By using the fact that $\left(AB{\right)}^{T}={B}^{T}{A}^{T}$ (where ${}^{T}$ denotes a transposed matrix), we get:
$\left[x{\right]}^{T}\left[{v}_{t}\right]\left[y\right]=\left(\left[T\right]\left[x\right]{\right)}^{T}\left[v\right]\left[y\right]=\left[x{\right]}^{T}\left[T{\right]}^{T}\left[v\right]\left[y\right]=\left[x{\right]}^{T}\left(\left[v{\right]}^{T}\left[T\right]{\right)}^{T}\left[y\right]$
So, that brings us to:
$\left[\xi {\right]}^{T}=\left[v{\right]}^{T}\left[T\right]$
Notice $\left[v\right]$ is invertible, as $v$ is non-degenerate, so:
$\left[T\right]=\left(\left[v{\right]}^{T}{\right)}^{-1}\left[\xi {\right]}^{T}$

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