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!
Question about Jordan canonical form watch
- Thread Starter
- 10-04-2011 22:26
- 10-04-2011 23:03
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.