$A$ is a $10X10$ matrix.

$$\operatorname{rank}\left(A^2\right)=2$$ $$\operatorname{rank}\left(A^3\right)=0$$

So from this I know that:

  1. All the eigenvalues are $0$ (so i'll just talk about the blocks's sizes)
  2. The minimal polynomial is $\lambda^3$, so the largest Jordan block is of size 3
  3. There are 2 blocks of at least size 3 (because of the nullity difference between the ranks) $\implies$ There are 3 blocks of size 3.

So I think the possible JCFs are:

  1. Two blocks of size 3, two blocks of size 2
  2. Two blocks of size 3, one block of size 2, two blocks of size 1
  3. Two blocks of size 3, four blocks of size 1

But I'm not sure how to find the dimension of the eigenspaces for $A^2,A^3$. Is it just the rank of the matrix itself?

  • 1
    $\begingroup$ The dimension of the Eigenspace for the eigenvalue $0$ is the dimension of the kernel, which by the dimension theorem is the dimension of the domain ($10$) minus the rank ($2$). $\endgroup$ – b00n heT Jul 2 '16 at 12:44
  • $\begingroup$ $A^2$ has rank 2, so the rank of $A$ is greater $\endgroup$ – InsideOut Jul 2 '16 at 12:55

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