One must find general solution for
$$y' = \left(\begin{matrix} 1&2&-1&-2&1&2\\ -1&-2&1&2&-1&-2\\ 2&4&-2&-4&2&4\\ -2&-4&2&4&-2&-4\\ 3&6&-3&-6&3&6\\ -3&-6&3&6&-3&-6 \end{matrix} \right)y$$
Well, I'm aware of the fact that the matrix is of rank 1. Nevertheless, I think I have to find eigenvalues and eigenvectors, so there must be a clever way to do that (which is, not involving Laplace expansion). Could you help me?
$\endgroup$ 01 Answer
$\begingroup$Since the matrix has rank 1, change to a basis where only the top row of the matrix has nonzero entries. Then $y_2$ through $y_6$ (in new coordinates) must clearly be constants but otherwise arbitrary. Now solve for $y_1$.
If you want to be slightly more systematic about it, it's easy enough to read off eigenvalues and eigenvectors. The eigenvector with nonzero eigenvalue is the one that generates the span, so $(1,-1,2,-2,3,-3)$, with eigenvalue (calculate directly) $-2$.
The null space has dimension 5, which gives us 5 independent eigenvectors with eigenvalue 0. For example, we can take
- $(-2,1,0,0,0,0)$
- $(+1,0,1,0,0,0)$
- $(+2,0,0,1,0,0)$
- $(-1,0,0,0,1,0)$
- $(-2,0,0,0,0,1)$
and these eigenvectors taken together diagonalize the matrix to $\Delta(-2,0,0,0,0,0)$.
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