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$A$ is a $10 \times 10$ nilpotent matrix of order $4$ ($A^4=0$) over $\mathbb C$ with $\operatorname{rank} (A)=6$.

Find all possible Jordan Canonical forms

The nullity of $A$ is $4$ so there are $4$ blocks. the rank is $6$ so the rank of the JCF matrix is also $6$. The minimal polynomial is $\lambda^4$ so there are $4$ blocks.

The options I figured are: $$\operatorname{diag}(J_{4}(0),J_{4}(0),J_1(0),J_1(0)),\\\operatorname{diag}(J_4(0),J_3(0),J_2(0),J_1(0)),\\\operatorname{diag}(J_4(0),J_2(0),J_2(0),J_2(0))$$

Is there anything i'm missing when it comes to analyzing the possible Jordan forms for nilpotent matrices?

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Your answer is correct, but this statement: "The minimal polynomial is $\lambda^4$ so there are 4 blocks." is incorrect...the minimal polynomial indicates the largest order of the Jordan blocks, so in this case we know that the largest order Jordan block is $J_4(0)$ - there must be at least one such block - that's all it tells you, not how many blocks there are.

The other information you have here is the rank of the matrix, so the sum of the ranks of the blocks must add up to 6. And you know the characteristic polynomial is $x^{10}$ - so the sum of the orders of the blocks must add up to 10.

Now what is the rank of the block $$J_4(0) = \begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \qquad?$$ It is rank 3 of course. So the remaining Jordan blocks must have rank 3 altogether.

So either we have 1 more block $J_4(0)$ and two blocks $J_1(0)$ (which has rank 0).

Or we have 1 block $J_3(0)$ which is rank 2, 1 block $J_2(0)$ which has rank 1, and 1 block $J_1(0)$ which has rank 0.

Lastly, you could have three blocks $J_2(0)$.

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  • $\begingroup$ Yes it's a typing mistake. Thank you! $\endgroup$ – PanthersFan92 Jul 4 '16 at 11:30

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