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January 31, 2015

January 31, 2015

Posted by **Christopher** on Thursday, July 12, 2012 at 8:30pm.

[3 2 4

2 0 2

4 2 3]

is distinguishable even though one eigenvector has algebraic multiplicity 2. Do this by brute force computation. Why would you expect this to be true, even without calculation?

Then, for the A, write A= Q lambda Q^(T) where Q's columns are orthogonal (unit) vectors of A.

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