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    Jordan Canonical Form

    Result

    (5 | 4 | 2 | 1
0 | 1 | -1 | -1
-1 | -1 | 3 | 0
1 | 1 | -1 | 2) = S.J.S^(-1)
where
S = (-1 | 1 | 1 | 1
1 | -1 | 0 | 0
0 | 0 | -1 | 0
0 | 1 | 1 | 0)
J = (1 | 0 | 0 | 0
0 | 2 | 0 | 0
0 | 0 | 4 | 1
0 | 0 | 0 | 4)
S^(-1) = (0 | 1 | 1 | 1
0 | 0 | 1 | 1
0 | 0 | -1 | 0
1 | 1 | 1 | 0)

    Back to List | POWERED BY THE WOLFRAM LANGUAGE