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Write down all the possible Jordan normal forms for matrices with characteristic polynomial $(x-a)^5$. In each case, calculate the minimal polynomial and the geometric multiplicity of the eigenvalue $a$. Verify that this information determines the Jordan normal form.

I found this question in a textbook that I'm using for a test I have tomorrow. I think that I need to use the method for finding the Jordan normal form of a matrix but I can't see how to apply it and I don't have much intuition about the answer...

I'm guessing that there are 5 possibilities since the minimal polynomial can be any factor of the characteristic, but I don't know how to prove this.

Some help would be great for my test tomorrow! Thanks

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Because there is only one eigenvalue, all the Jordan blocks in the JNF will have eigenvalue $a$, and so the only question is the sizes of those blocks. Note that the ordder of the blocks does not matter, as putting the blocks in a different order gives a similar matrix.

The minimal polynomial gives you the size of the largest block (why?), and the geometric multiplicity of the eigenvalues gives you the number of Jordan blocks, as each block will have a one dimensional eigenspace$.^{\dagger}$ Since the characteristic polynomial is $(x-a)^5$, the sum of the sizes of the blocks must be $5$. We therefore have that the collection of possibly Jordan normal forms is in bijection with ways of writing $5$ as the sum positive integers where we don't care about the order [and so might as well put the larger numbers at the front], and we wish to show that knowing the sum, the largest summand, and the number of summands determines in the sum when the sum is $5$. (it will NOT be true for large numbers. For example, since $3+3+1=3+2+2$, the statement is false if we replaced $5$ with $7$).

Can you take it from here?

$^{\dagger}$Previously, I stated that the geometric multiplicity was the number of blocks of size one, which was incorrect.

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  • $\begingroup$ So, are there 15 possibilities? $\endgroup$ – derivative Nov 21 '13 at 16:45
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    $\begingroup$ I count $7$ ways: $[5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1], [1,1,1,1,1]$ $\endgroup$ – BaronVT Nov 21 '13 at 17:07

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