How can I determinate the bases for the most simple representation of a linear transformation? Imagine a linear transformation $\Phi : \mathbb{R}^4 \rightarrow \mathbb{R}^3$ with the ordered standard basis:
$B = (\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix})$ and 
$C = (\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix})$. The transformation is defined by $M^B_C (\Phi) = \begin{pmatrix} 0 & 0 & 0 & 0 \\1 & 2 & 3 & 4 \\5 & 6 & 7 & 8 \end{pmatrix}$
Lets define some other basis: 
$D = (\begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}, \begin{pmatrix} 1 \\ 3 \\ 3 \\ 7 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 4 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 7 \\ 1 \\ 8 \end{pmatrix})$ and $E = (\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ 5 \\ 7 \end{pmatrix}, \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix})$
In this case you get this transformation matrix:
$M^D_E (\Phi) = 
\begin{pmatrix} 
110 & 156 &  93 & 195 \\ 
10  & 16  &   3 &  15 \\ 
-50 & -72 & -39 & -87 
\end{pmatrix}$
If you have $M^D_E(\Phi)$ given and you want to find the two Basis B and C so that the linear transformation $M^B_C(\Phi)$ is as simple as possible. How would you do that?
(simple means as many zeros as possible)
I guess one step could be to determine the null space.
$\left \{ \begin{pmatrix}-39\\ 15\\ 0\\ 10\end{pmatrix}, \begin{pmatrix}-51\\ 30\\ 10\\ 0\end{pmatrix} \right \}$

Try the singular value decomposition, which gives orthonormal bases in the domain and in the codomain with respect to which the linear transformation is given by a diagonal matrix.
