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Prove that characteristic polynomial of a complex matrix $A$ divides its minimal polynomial if and only if all eigenspaces of $A$ are one-dimensional.

As far as I can see I the only possible case is when minimal polynomial equals characteristic one.

All distinct eigenvalues with multiplicity 1 grant us that the eigenspaces would be one-dimensional, I thought that this is the key to the solution, however we have the theorem stating that on the other side, eigenspace dimension could be less then or equal to its eigenvalue algebraic multiplicity. Everything now mixed up, will be thankful for any help.

Zev Chonoles
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mouse
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    Since the minimal polynomial always divides the charateristic polynomial and they are both monic, this is equivalent to this thread. See in particular Arturo Magidin's answer. –  Aug 14 '13 at 01:56

1 Answers1

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Hint:

They have the same roots, and $m_A(x)\mid p_A(x)$, so you're just concerned when the roots have the same multiplicity.

Note then that if $A$ is a complex matrix, then the degree of $x-\lambda$ in $m_A(x)$ is size of the largest Jordan block corresponding to $\lambda$, the degree of $x-\lambda$ in $p_A(x)$ is the sum of the dimensions of all the Jordan blocks associated to $\lambda$, and the geometric multiplicity is the total number of Jordan blocks associated to $\lambda$.

Alex Youcis
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  • I don't quite understand why for a complex matrix the degree of x−λ in mA(x) is size of the largest Jordan block corresponding to λ while the degree of x−λ in pA(x) is the sum of blocks dimensions? Could you please explain it a bit more detailed? – mouse Aug 13 '13 at 22:45
  • Well, the second question is easy because a Jordan matrix is upper triangular, and so the algebraic multiplicity is literally just the number of times that eigenvalue appears along the diagonal, which is precisely the sum of the dimensions of the Jordan blocks associated to that eigenvalue. For the first, think about what $J-\lambda I$ to a given power looks like. Convince yourself that the size of the largest Jordan block associate to $\lambda$ is the minimal power such that all the Jordan blocks associated $\lambda$ vanish. – Alex Youcis Aug 13 '13 at 22:54
  • Now, imagine doing this as you take the product over all $A-\lambda I$ for all eigenvalues $\lambda$, remembering to multiply block diagonally. You should then be able to convince yourself that the minimal degree polynomial which annihilates the Jordan matrix is $(x-\lambda_1)^{m_1}\cdots(x-\lambda_n)^{m_n}$ where $m_i$ is the size of the largest Jordan block. @mouse – Alex Youcis Aug 13 '13 at 22:55
  • Thank you, it seems clear now. However I still have no idea how to connect the size of the blocks in Jordan matrix with the condition that all of the eigenspaces should be one-dimensional – mouse Aug 13 '13 at 23:17
  • @mouse If you have more than one Jordan block, you're totally screwed--the sum of the dimensions of the Jordan blocks (the multiplicity of $\lambda$ in $p_A(X)$ will necessarily be greater than the size of the largest Jordan block (the multiplicity of $\lambda$ in $m_A(X)$. So, you can only have on Jordan block...so. – Alex Youcis Aug 13 '13 at 23:20
  • ... so that means that I have only distinct eigenvalues and, and thus only one-dimensional eigenspaces? – mouse Aug 13 '13 at 23:35
  • @mouse So, that means you can only have one Jordan block. Since the geometric multiplicity is number of Jordan blocks... – Alex Youcis Aug 13 '13 at 23:36