3
$\begingroup$

Related to this question.

I am trying to understand how to show all nilpotent matrices of size $4\times 4$ using the Jordan normal form.

From what I understand, we string together Jordan blocks of different sizes. I am trying to find how to find all forms of some matrix $A$ where $A$ is a $4\times 4$ matrix, for $A^2=0$ using these blocks.


Since all the eigenvalues of a nilpotent matrix are $0$, the Jordan form should look like this I thought:

$$\def\b{\begin{bmatrix}}\def\e{\end{bmatrix}} \b \lambda&1&0&0\\0&\lambda&1&0\\0&0&\lambda&1\\0&0&0&\lambda\e=\b 0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\e$$ As a single Jordan block(which seems right, since I have one eigenvalue $0$)

This isn't nilpotent of degree $2$

Also Will Jagy suggested in my previous question that I use $2$ $2\times 2$ matrices, or a $2\times 2$ and two $1\times 1$ blocks. I can see that two $2\times 2$ blocks do give me $A^2=0$, but I can't get this to work with one $2\times 2$ block and two $1\times 1$ blocks.

$\endgroup$

1 Answer 1

3
$\begingroup$

$$ A = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) $$ is, in your words, nilpotent "of degree 4" because $A^4 = 0$ but not lower. This is the same as saying that the characteristic polynomial and the minimal polynomial are both $A^4.$ This is a single Jordan block, 4 by 4. The condition that characteristic and minimal polynomials agree is precisely that every eigenvalue occur in just one Jordan block; we sometimes need to extend to the complex numbers to have any eigenvalues at all.

Next, $$ B = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) $$ is nilpotent of degree 3, minimal poly is $B^3.$ There is one 3 by 3 block in the upper left, then a 1 by 1 block (no off-main diagonal anything). It might help you visulize this if you replace the main diagonal with the same entry, say $\lambda = 5,$ doesn't matter. It then becomes easy to see how the final $5$ is in a little 1 by 1 block all alone.

$$ C = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) $$ has minimal polynomial $C^2,$ two 2 by 2 blocks.

$$ D = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) $$ has minimal polynomial $D^2,$ one 2 by 2 block and two 1 by 1 blocks (try putting a $5$ on the main diagonal).

$$ E = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) $$ has minimal polynomial $E,$ four 1 by 1 blocks (try putting a $5$ on the main diagonal). That is, it is already the zero matrix.

WHY NOT? REPEATING WITH ALL MAIN DIAGONAL ELEMENTS EQUAL TO $5.$ I put in lines to emphasize, only the square blocks along the main diagonal matter; it would be much more work to make a little square around each block, but not hard when I let the lines go all the way through.

$$ A = \left( \begin{array}{cccc} 5 & 1 & 0 & 0 \\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 5 \end{array} \right) $$ This is the same as saying that the characteristic polynomial and the minimal polynomial are both $(A-5)^4.$ This is a single Jordan block, 4 by 4. The condition that characteristic and minimal polynomials agree is precisely that every eigenvalue occur in just one Jordan block; we sometimes need to extend to the complex numbers to have any eigenvalues at all.

Next, $$ B = \left( \begin{array}{ccc|c} 5 & 1 & 0 & 0 \\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 0 \\ \hline 0 & 0 & 0 & 5 \end{array} \right) $$ minimal poly is $(B-5)^3.$ There is one 3 by 3 block in the upper left, then a 1 by 1 block (no off-main diagonal anything).

$$ C = \left( \begin{array}{cc|cc} 5 & 1 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ \hline 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 5 \end{array} \right). $$ $C$ has minimal polynomial $(C-5)^2,$ two 2 by 2 blocks.

$$ D = \left( \begin{array}{cc|c|c} 5 & 1 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ \hline 0 & 0 & 5 & 0 \\ \hline 0 & 0 & 0 & 5 \end{array} \right). $$ $D$ has minimal polynomial $(D-5)^2,$ one 2 by 2 block and two 1 by 1 blocks.

$$ E = \left( \begin{array}{c|c|c|c} 5 & 0 & 0 & 0 \\ \hline 0 & 5 & 0 & 0 \\ \hline 0 & 0 & 5 & 0 \\ \hline 0 & 0 & 0 & 5 \end{array} \right). $$ $E$ has minimal polynomial $E-5,$ four 1 by 1 blocks.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much, this along with some chat on the M.SE chat page has cleared up my confusion. I appreciate the large time this could have taken! $\endgroup$
    – user142198
    Commented Mar 12, 2015 at 2:48

You must log in to answer this question.