show that $[f]_{B_1}$ is congruent to $[f]_{B_2}$

Question

Let $f$ be a bilinear from on $\mathbb{R}^2$ defined by $$f(u,v)=2u_1v_1-3u_1v_2+u_2v_2,\quad u=(u_1,u_2),v=(v_1,v_2)$$ $(i)$ Find the matrices $[f]_{B_1}$ and $[f]_{B_2}$ relative to the bases $B_1=\{(1,0),(1,1)\}$ and $B_2=\{(2,1),(1,-1)\}$, respectively.
$(ii)$ Hence show that $[f]_{B_1}$ is congruent to $[f]_{B_2}$.

I manage to do $(i)$. And find the matrices $$[f]_{B_1}=\begin{pmatrix}2 &-1\\2 &0 \end{pmatrix} \qquad [f]_{B_2}=\begin{pmatrix}3 &9\\0 &6 \end{pmatrix}$$ But seems those matrices are not symmetric I don't know how to get any decomposition $[f]_{B_1}=P^T[f]_{B_2}P($in order to show congruent$)$.
Any help will be appreciated.
Thanks in advance

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Using @Trevor Gunn comment,
$$[f]_{B_2}=\underbrace{\begin{pmatrix}2&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\-1&1\end{pmatrix}}_{\begin{pmatrix}1&1\\ 2&-1\end{pmatrix}=P^T}[f]_{B_1}\underbrace{\begin{pmatrix}1&-1\\0&1\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix}}_{\begin{pmatrix}1&2\\1&-1\end{pmatrix}=P}$$ it seems lot of calculation needed to find those $P^T$ and $P$. Any easier approach do exist$?$

Answer 1

Let T be a linear operator such that $T(u_i)=v_i$. The matrix of $T$ relative to the basis $B_1$, $$T(1,0)=(2,1)=(1)(1,0)+(1)(1,1)$$ $$T(1,1)=(1,-1)=(2)(1,0)+(-1)(1,1)$$ Therefore, coefficient matrix of $T$ relative to the basis $B_1$ is $\begin{pmatrix}1&1\\2&-1\end{pmatrix}$ Now the transition matrix $P$, $$\begin{pmatrix}1&1\\2&-1\end{pmatrix}^T=\begin{pmatrix}1&2\\1&-1\end{pmatrix}=P$$ Can you go from here$?$

Answer 2

The matrices are not symmetric because $f$ is not symmetric.

For (ii), it is perhaps helpful to think about $[f]_B$ where $B$ is the standard basis. We have

\begin{align} \begin{pmatrix} 1 & 0 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= 2,& \begin{pmatrix} 1 & 0 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 1 \end{pmatrix} &= -1 \\ \begin{pmatrix} 1 & 1 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= 2,& \begin{pmatrix} 1 & 1 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 1 \end{pmatrix} &= 0. \end{align}

So maybe you can see why

$$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^\top [f]_B \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = [f]_{B_1}. $$

If you can figure this out, you should be able to finish (ii).