$V:= \begin{pmatrix} a&b&c&d\\b&a&d&c\\c&d&a&b\\d&c&b&a\\\end{pmatrix}$ $W:= \begin{pmatrix} u&v&w\\x&y&0\\z&0&0\\\end{pmatrix}$ $\phi: V \rightarrow W, \begin{pmatrix} a&b&c&d\\b&a&d&c\\c&d&a&b\\d&c&b&a\\\end{pmatrix} \mapsto \begin{pmatrix} a-b&a+c&a-d\\b+c&b-d&0\\c+d&0&0\\\end{pmatrix}$
So $D^B_C(\phi)$ is bascially a transformation of the spanvectors of V to W. Now the basis of $V$ are obviously $B: \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}, \begin{pmatrix} 0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\\\end{pmatrix}, \begin{pmatrix} 0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\\\end{pmatrix}, \begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\\\end{pmatrix}$ and the basis of $W$ $C: \begin{pmatrix} 1&0&0\\0&0&0\\0&0&0\\\end{pmatrix}, \begin{pmatrix} 0&1&0\\0&0&0\\0&0&0\\\end{pmatrix}, \begin{pmatrix} 0&0&1\\0&0&0\\0&0&0\\\end{pmatrix}, \begin{pmatrix} 0&0&0\\1&0&0\\0&0&0\\\end{pmatrix}, \begin{pmatrix} 0&0&0\\0&1&0\\0&0&0\\\end{pmatrix}, \begin{pmatrix} 0&0&0\\0&0&0\\1&0&0\\\end{pmatrix}$
Now I have no idea how I get the transformation matrix, help is very appreciated