Let $A=\begin{bmatrix}1 & 2\\-1 & 1\end{bmatrix}$. Find a Matrix $B$ s.t for any $u,v \in \Bbb{R}^2, (u,Av)=(Bu,v)$

Question

Let $A=\begin{bmatrix}1 & 2\\-1 & 1\end{bmatrix}$. Find a $B\in \Bbb{M}_2(\Bbb{R})$ such that for any $\textbf{u,v} \in \Bbb{R}^2, (\textbf{u},A\textbf{v})=(B\textbf{u},\textbf{v})$ or prove that no such $B$ exists.

Answer 1

The most basic and mechanical way to do this is by computing the products explicitly. $$ <u,Av> = u_1v_1 + 2u_1v_2 - u_2v_1 + u_2v_2 $$ $$ <Bu,v> = b_{11}u_1v_1 + b_{21}u_1v_2 + b_{12}u_2v_1 + b_{22}u_2v_2 $$ and you get $b_{11}=1; b_{12}=-1; b_{21}=2 ; b_{22}=1$, that is, $B=A^T$, as a comment suggests.

Answer 2

If you're considering the standard inner product, namely $(u,v)=u^Tv$, then you want to ensure that $$ u^TAv=(Bu)^Tv $$ The right hand side becomes $u^TB^Tv$, so your equality is $$ u^TAv=u^TB^Tv $$ Can you spot that $B^T=A$ is a solution? And the only one, by the way, because the equality can be written $$ u^T(A-B^T)v=0 $$ for every $u,v$. If $X$ is a nonzero matrix, there is $v$ such that $Xv\ne0$. Then there exists $u$ such that $u^TXv\ne0$.