Set M to be the set of all matrices of the form: $\bigl( \begin{smallmatrix} 0 & b \\ c & 0 \end{smallmatrix} \bigr)$, with b, c being real numbers.
A basis for M (given) is $\mathcal E$ = $\{[\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}] ,[\begin{smallmatrix} 0 & 0 \\ 1 & 0 \end{smallmatrix}] \}$,
with the elements referred to as $\mathcal E = \{e_1, e_2\}$
Another basis for M (given) is $\mathcal B$ = $\{[\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}] ,[\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}] \}$, with the elements referred to as $\mathcal B = \{b_1, b_2\}$
Find the change of bases matrix from basis $\mathcal B$ to basis $\mathcal E$.
I'm not really sure how to approach this problem, as I've never seen matrices being used as parts of a basis. I read on this question that I can think of these as 4-tuples, but then the matrices $[e1\;e2]$ and $[b1\;b2]$ aren't square, so I can't do the row reduction method.
I see that $e_1 = b_1 + b_2$ but I don't see how to obtain $e_2$.
Best Answer
If $x=\beta_1\,b_1+\beta_2\,b_2$, and $$ b_1=a_{11}\,e_1+a_{21}\,e_2,\ \ \ b_2=a_{12}\,e_1+a_{22}\,e_2, $$ then $$ x=(a_{11}\beta_1+a_{12}\beta_2)\,e_1+(a_{21}\beta_1+a_{22}\beta_2)\,e_2, $$ and so the transformation matrix is $$ A=\begin{bmatrix}a_{11}& a_{12}\\ a_{21}& a_{22}\end{bmatrix}.$$ Indeed, $$ \begin{bmatrix}a_{11}& a_{12}\\ a_{21}& a_{22}\end{bmatrix}\,\begin{bmatrix}\beta_1\\ \beta_2\end{bmatrix}=\begin{bmatrix}a_{11}\beta_1+a_{12}\beta_2 \\ a_{21}\beta_1+a_{22}\beta_2\end{bmatrix}. $$
In this concrete example, $$ b_1=e_1+e_2,\ \ b_2=e_1-e_2. $$ So the change of bases matrix is $$ A=\begin{bmatrix}1&1\\ 1&-1\end{bmatrix}. $$