$A$ is matrix under $R$ which I know the following information about it:

$f_A(x)=(x+2)^4x^4$- Characteristic polynomial

$m_A(x)=(x+2)^2x^2$- Minimal polynomial.

I'm trying to find out

(i) $A$'s rank

(ii) $\dim$ $\ker(A+2I)^2$

(iii) $\dim$ $\ker (A+2I)^4$

(iv) the characteristic polynomial of $B=A^2-4A+3I$.

I believe that I don't have enough information to determine none of the above.

By the power of $x$ in the minimal polynomial I know that the biggest Jordan block of eigenvalue 0 is of size 2, so there can be two options of Jordan form for this eigenvalue: $(J_2(0),J_2(0))$ or $(J_2(0),J_1(0),J_1(0))$, therefore $A$'s rank can be $2$ or $3$. I'm wrong, please correct me.

How can I compute the rest?

Thanks for the answers.

  • $\begingroup$ Hint: For a root of the characteristic polynomial $r$, if its multiplicity in the characteristic polynomial is $n$ then its multiplicity in the minimal polynomial is the smallest k such that $\dim\ker((A-rI)^k)=n$. $\endgroup$ – user13838 Aug 21 '11 at 10:38
  • $\begingroup$ You mean that $\dim$ $\ker(A+2I)^2$=$\dim$ $\ker (A+2I)^4$? Still Do I have enought information tell what is $\ker(A+2I)^2$? $\endgroup$ – Jozef Aug 21 '11 at 10:43
  • $\begingroup$ Yes that's correct. You do have some information coming from the fact that when you square a Jordan block in the form of $(\lambda I -A)$ it's rank drops by one. Or pictorially the 1's on the superdiagonal shifts towards the upper right corner, right? Here, after squaring, the dimension of the kernel stays the same. Then, this should give us an idea about the size of the Jordan blocks. $\endgroup$ – user13838 Aug 21 '11 at 10:52
  • $\begingroup$ Hint regarding the rank: the behaviour of the generalized eigenspace for $-2$ is just the same as the behaviour on the generalized eigenspace for $0$ $\endgroup$ – Geoff Robinson Aug 21 '11 at 14:31
  • $\begingroup$ Aren't there two options for (i)? I still don't manage to understand if I'm wrong or right. $\endgroup$ – Jozef Aug 21 '11 at 14:57

If the Jordan form of A is C then let P be invertible such that $A=PCP^{-1}$ then $$(A+2I)^2=P(C+2I)^2P^{-1}\;\rightarrow\; \dim\ker((A+2I)^2)=\dim\ker((C+2I)^2)$$ and you know exactly how $(C+2I)^2$ looks like (well, at least the part of the kernel).

The same operation should help you solve the rest of the problems

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  • $\begingroup$ I still don't understand how should I compute rank $A$?... Is there one options of this rank? I think that it can either 2 or 3. Is this answer only (ii)+(iii)? $\endgroup$ – Jozef Aug 21 '11 at 13:36
  • $\begingroup$ For example, can you see the rank of the following Jordan block structure? $$ (\lambda I - J) = \begin{pmatrix} 0&1&0\\0&0&1\\0 &0&0\end{pmatrix}$$. Compute the square of this and check the rank of the result. This will give you an idea of rank change and the structure of the Jordan block. Note that the rank of the matrix is related to the number of the zero eigenvalues. $\endgroup$ – user13838 Aug 21 '11 at 19:25

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