Consider $g : M (R)_{2x2}$ → $R$ given by $g(A)=a_{11} + 2a_{12} + 3a_{32} +4a_{22}$. We consider on $M_{2x2} (R)$ the inner product given by $<A,B> = tr(A^t ,B)$. Find the vector $y$.
I only could get the orthonormal basis (obviously), but I don't know exactly where to go from there. Should I simply apply $\sum$ $\overline{g(e_i)}$ $e_i$? I'm having trouble figuring out the computation method.
This is the orthonormal basis:
Let β = $$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}, $$
$$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}, $$
$$ \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}, $$
$$ \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}, $$
In your case the 'vector' $y$ is the matrix $B$ in the vectorspace $M_{2\times 2}(R)$. Assume the matrices are given by $$ A= \left(\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}\right) , \quad B = \left(\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}\right), $$
then the scalar product is given by \begin{align} \langle A, B\rangle &= \text{tr} (A^t B ) = \text{tr}\left(\begin{matrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{matrix}\right) \cdot \left(\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}\right) = \text{tr} \left(\begin{matrix} a_{11}b_{11}+a_{21}b_{21} & a_{11}b_{12}+a_{21}b_{22} \\ a_{12}b_{11}+a_{22}b_{21} & a_{12}b_{12}+a_{22}b_{22} \end{matrix}\right) \\ &= a_{11}b_{11}+a_{21}b_{21} + a_{12}b_{12}+a_{22}b_{22} \stackrel{!}{=} g(A), \end{align} so if $g(A)=a_{11}+2 a_{12}+3 a_{21} + 4 a_{22}$, then the matrix you search is $$ B=\left(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}\right) . $$