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Suppose $J$ is an $n \times n$ Jordan block with eigenvalue $0$, what is the Jordan form of $J^2$?

My solution:

I squared the matrix, it follows that the eigenvalues are $0$ again, and $\dim(N(J^2)) = 2$, with eigenvectors $e_{n-1}$, $e_{n}$.

Therefore, the Jordan form of $J^2$ is an $(n-2) \times (n-2)$ block and two $1\times1$ blocks all with eigenvalue zero.

However, there is a hint with the question which says to treat odd and even values of $n$ separately but I fail to see why.

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You're wrong. For example, $$ \pmatrix{0 & 1 & 0 \cr 0 & 0 & 1\cr 0 & 0 & 0\cr}^2 = \pmatrix{0 & 0 & 1\cr 0 & 0 & 0\cr 0 & 0 & 0\cr} \ \text{has Jordan form}\ \pmatrix{0 & 1 & 0\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$$ (a block of size $2$ and a block of size $1$) but $$ \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr}^2 = \pmatrix{0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr}\ \text{has Jordan form}\ \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr}$$ (two blocks of size $2$).

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Let me illustrate this with the $n = 6$ case. Here, we have a basis $\{ e_1, e_2, e_3, e_4, e_5, e_6 \}$, and the action of $J$ on the basis vectors is given by $$ Je_1 = e_2, \ \ Je_2 = e_3, \ \ Je_3 = e_4, \ \ Je_4 = e_5, \ \ Je_5 = e_6, \ \ Je_6 = 0. $$ So the action of $J^2$ on this basis is given by

$$ J^2e_1 = e_3, \ \ J^2e_2 = e_4, \ \ J^2e_3 = e_5, \ \ J^2e_4 = e_6, \ \ J^2e_5 = 0, \ \ J^2e_6 = 0.$$

[You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]

But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:

  • There is the subspace spanned by $\{ e_1, e_3, e_5 \}$, on which the action of $J^2$ is $$ J^2e_1 = e_3, \ \ J^2e_3 = e_5, \ \ J^2e_5 = 0.$$
  • Then there is the subspace spanned by $\{ e_2, e_4, e_6 \}$, on which the action of $J^2$ is $$ J^2e_2 = e_4, \ \ J^2e_4 = e_6, \ \ J^2e_6 = 0 .$$

So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is $$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$$

That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?

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  • $\begingroup$ Thank you that clears things up I can finish it out from there. More generally when does the "key point" come into play? and do you have any references I could see about that because I've always assumed if you have eigenvectors for an eigenvalue of multiplicity 'n' it follows that you get an (n - 2) jordan block automatically $\endgroup$ – Scosh_lr Dec 25 '18 at 22:32
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    $\begingroup$ @Scosh_lr I'm afraid I don't think I've ever come across such a theorem. To give you another example, suppose we have a $4\times 4$ matrix $A$, and I tell you that $A$ has $0$ as an eigenvalue, and no other eigenvalues, and that the dimension of the eigenspace corresponding to the zero eigenvalue is 2-dimensional. Then there are two possible Jordan forms for $A$: either it has a 3-dimensional block and a 1-dimensional block, or it has two 2-dimensional blocks. The fact that the dimension of the eigenspace is two means that we have two blocks. $\endgroup$ – Kenny Wong Dec 25 '18 at 22:43
  • $\begingroup$ why is it not possible to have a 2x2 block and two 1x1 blocks? $\endgroup$ – Scosh_lr Dec 25 '18 at 22:57
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    $\begingroup$ @Scosh_lr Because in that case, the eigenspace would have dimension three. (To spell it out, if $e_1, e_2$ span the $2\times 2$ block, and $e_3$ spans a $1 \times 1 $ block and $e_4$ spans the other $1 \times 1$ block, then $Ae_1 = e_2$, $Ae_2 = 0$, $Ae_3 = 0$ and $Ae_4 = 0$, so $e_2, e_3, e_4$ are all eigenvectors with eigenvalue $0$.) $\endgroup$ – Kenny Wong Dec 25 '18 at 23:24
  • $\begingroup$ Oh I see now, Thanks for the help. $\endgroup$ – Scosh_lr Dec 25 '18 at 23:26

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