Calculate $D_E^E$ and $D_B^B$ from $\Phi: \mathbb{R^5} \rightarrow \mathbb{R^5} (x1,x2,x3,x4,x5)' \mapsto (0,x2,0,x4,0)'$

Question

Calculate $D_E^E$ and $D_B^B$ from $\Phi: \mathbb{R^5} \rightarrow \mathbb{R^5} (x1,x2,x3,x4,x5)' \mapsto (0,x2,0,x4,0)'$

E: $\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\\0 \end{pmatrix},\begin{pmatrix}0\\0\\1\\0\\0 \end{pmatrix},\begin{pmatrix}0\\0\\0\\1\\0 \end{pmatrix},\begin{pmatrix}0\\0\\0\\0\\1 \end{pmatrix}$ B: $\begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix},\begin{pmatrix}1\\1\\1\\1\\0 \end{pmatrix},\begin{pmatrix}1\\1\\1\\0\\0 \end{pmatrix},\begin{pmatrix}1\\1\\0\\0\\0 \end{pmatrix},\begin{pmatrix}1\\0\\0\\0\\0 \end{pmatrix}$

For $D^E_E$, if one takes only the $e^2,e^4$ then it'd work $A :=\begin{pmatrix}0&0&0&0&0\\0&1&0&0&0\\0&0&0&0&0\\0&0&0&1&0\\0&0&0&0&0 \end{pmatrix} \cdot \begin{pmatrix}a\\b\\c\\d\\e \end{pmatrix} = \begin{pmatrix}0\\b\\0\\d\\0 \end{pmatrix}$ now im not too sure how to calculate it or if that is even allowed.

Answer 1

The matrix you've written is indeed $D_E^E$.

We can calculate $D_B^B$ column by column. For example: $$ \Phi(b^2) = \pmatrix{0\\1\\0\\1\\0} = (0)b^1 + (1)b^2 + (-1)b^3 + (1)b^4 + (-1)b^5 $$ So, the second column of $D_B^B$ is given by $(0,1,-1,1,-1)'$. Do this for each column to find $$ D_B^B = \pmatrix{ 0&0&0&0&0\\1&1&0&0&0\\-1&-1&0&0&0\\ 1&1&1&1&0\\-1&-1&-1&-1&0} $$