Dania Mueller

2022-06-27

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 $\xi ={v}_{T}$.

Assuming $v$ is nondegenerate, let us have another bilinear form $\xi \in {T}_{2}(V)$. Prove that there exists exactly one transformation $T$ so $\xi ={v}_{T}$.

Tianna Deleon

Beginner2022-06-28Added 29 answers

By using the fact that $(AB{)}^{T}={B}^{T}{A}^{T}$ (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:

$[T]=([v{]}^{T}{)}^{-1}[\xi {]}^{T}$

$[x{]}^{T}[{v}_{t}][y]=([T][x]{)}^{T}[v][y]=[x{]}^{T}[T{]}^{T}[v][y]=[x{]}^{T}([v{]}^{T}[T]{)}^{T}[y]$

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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