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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. $\left \{ \begin{pmatrix}-39\\ 15\\ 0\\ 10\end{pmatrix}, \begin{pmatrix}-51\\ 30\\ 10\\ 0\end{pmatrix} \right \}$

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    $\begingroup$ Pick a basis $e_1, e_2, e_3, e_4$ of $\mathbb{R}^4$ for the domain and complete some linearly independent subset of $\Phi e_1, \Phi e_2, \Phi e_3, \Phi e_4$ to a basis of the range... $\endgroup$ Mar 23, 2012 at 1:17

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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.

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  • $\begingroup$ I can get these with Wolfram|Alpha: wolframalpha.com/input/… - so $\Sigma$ would be my new transformation matrix and U and V would be the new basis? Is there a way to get a representation with integers only? $\endgroup$ Mar 23, 2012 at 1:33
  • $\begingroup$ @moose, no, because $U$ and $V$ are orthogonal matrices. But you can do so if you accept an scalar multiple of an integer matrix. Take for instance the standard basis in $\mathbb R^2$ rotated 45 degrees or $\{(3,4), (-4,3)\}$ normalized. Of course, in higher dimension it's not so easy but on the other hand you have more room. $\endgroup$
    – lhf
    Mar 23, 2012 at 1:47

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