Characteristic and minimal polynomial properties

**davesface** · Dec 9th 2010, 05:36 PM

Consider a vector space V over $\mathbb{C}$ mapped to itself by the operator L. The characteristic polynomial is $c(\lambda)=(\lambda-1)^4(\lambda-2)^4(\lambda-3)^4$ and the minimal polynomial is $m(\lambda)=(\lambda-1)^3(\lambda-2)^2(\lambda-3)$ .

Determine whether or not each of the following properties can be determined uniquely from these 2 polynomials. If so, state it and argue why it is unique. If not, give a counterexample of distinct operators with the given minimal and characteristic polynomials:

a. dimension of V

c. geometric multiplicity of $\lambda$ =1, 2, and 3
d. the number of linearly independent eigenvectors for L.

**dwsmith** · Dec 9th 2010, 05:41 PM

No b?
a: Not sure.

c: 4 for 1 since you have $\displaystyle (\lambda-1)^4=(\lambda-1)(\lambda-1)(\lambda-1)(\lambda-1)$
4 for 2
3 for 3

d: can't be determined, I think.

**davesface** · Dec 9th 2010, 05:47 PM

Originally Posted by dwsmith

No b?
a: Not sure.

c: 4 for 1 since you have $\displaystyle (\lambda-1)^4=(\lambda-1)(\lambda-1)(\lambda-1)(\lambda-1)$
4 for 2
3 for 3

d: can't be determined

I fixed the characteristic polynomial, that was just a typo. Each should be to the 4th power. In any case, isn't that just the algebraic multiplicity?

b just asked for the eigenvalues of L, which are obvious.

**FernandoRevilla** · Dec 9th 2010, 05:47 PM

Clearly,

$\dim V=4+4+3=11$

For the other questions, use possible canonic forms of Jordan.

Regards.

Fernando Revilla

Edited: I didn't see the previous post, then $\dim V=12$ .

**dwsmith** · Dec 9th 2010, 05:49 PM

You are correct. We need to know the eigenspace for each eigenvalue to have the geo multiplicity.

**davesface** · Dec 9th 2010, 06:23 PM

Is it possible to say anything about the eigenspaces without knowing L? It would seem to me like the dimension of the eigenspace of each eigenvalue would be its exponent in the minimal polynomial since that is required to annihilate the matrix representation of L (GM of 1 is 3, GM of 2 is 2, GM of 3 is 1), but I can't find anything that would support this hunch.

As for part d, it seems like the way to go would be to just say that the number of linearly independent eigenvectors is the same as the number of blocks in the Jordan canonical form, but again I don't see how we can say anything about that without knowing either L or the geometric multiplicities of each eigenvalue (in which case the answer would be the sum of the GMs I believe).

**dwsmith** · Dec 9th 2010, 06:25 PM

I don't think we can say anything about the eigenspace since a multiplicity n of an eigenvalue doesn't guarantee that there will be n eigenvectors.

**davesface** · Dec 9th 2010, 08:05 PM

Well, I'm totally stumped on it. It just seems like there isn't enough information to answer c or d, but at the same time it seems like it would be almost impossible to come up with distinct operators with the same characteristic and minimal polynomials to use as a counterexample. Anyways, thanks for the help.

**TheEmptySet** · Dec 9th 2010, 08:36 PM

The muliplicity of each factor in the minimum polynomial tells you the size of the largest Jordan block corresponding to that eigenvalue.

So for $\lamba=1$ the largest Jordan block is 3x3 so the other must be a 1x1 block. So That eigenspace has two linearly independent eigenvectors.

For $\lambda =2$ the largest Jordan block is 2x2. Now this leaves two possibleities the other could be another 2x2 or 2 1x1's.

For $\lambda =1$ the largest Jordan block is 1x1. This gives 4 1x1 blocks, so you have 4 linearly independent eigenvectors.

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