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    Question 11(ii).

    I see that, by the JCF theorem, it's enough to show that a Jordan block J_n(x) has a square root. But how would I find the square root of this matrix? Obviously the leading diagonal will have x^0.5, but what would I choose for the rest of the entries? I couldn't even do it in the 2x2 case!

    I would take an arbitrary Jordan block, calculate the square, and then "reverse engineer" what the square root must be. (Doing a couple of concrete examples for the 2x2 case might help). I don't think it's very difficult.
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Updated: April 10, 2011
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