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$\def\b{\begin{bmatrix}}\def\e{\end{bmatrix}}$

Are $\b0&0&a&b\\0&0&c&d\\0&0&0&0\\0&0&0&0\e$ and it's transpose, $a,b,c,d\in \Bbb C$ the only nilpotent degree $2$ 'families' of matrices of size $4\times 4$?

I believe they are, but I wanted to verify.

  • it is spelled nilpotent. What does "degree 2" mean? – Will Jagy Mar 11 '15 at 03:32
  • @WillJagy $A^n=0$ is nilpotent of degree $n$ I was told earlier today –  Mar 11 '15 at 03:33
  • @WillJagy So here $A^2 = 0$ and I believe the matrices are the only matrices with (above as $A$) $A^2=0$ –  Mar 11 '15 at 03:38
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    Alright, but given any invertible matrix $P,$ we get another matrix with the same characteristic and minimal polynomials from $P^{-1} A P.$ This other matrix may have quite a different appearance. You might experiment with $P$ a permutation matrix, each row and each column have only a single $1,$ all other entries are $0.$ Not too difficult to find $P^{-1}$ – Will Jagy Mar 11 '15 at 03:42
  • @WillJagy So looking at similar matrices to this one, I will find ones of a different form is what you are suggesting(or I have misinterpreted) –  Mar 11 '15 at 03:44
  • Correct.............. You will see this in a different way once you get to Jordan Normal Form. The overall answer is the same, though, all matrices with a given minimal polynomial $A^2$ of a fixed size are equivalent to certain Jordan forms. – Will Jagy Mar 11 '15 at 03:46
  • @WillJagy Thank you that makes sense, I will investigate –  Mar 11 '15 at 03:48
  • @WillJagy Is it possible you could give me a hint on what 'certain Jordan forms' you refer? The only eigenvalue is $0$, so we have $\lambda^n=0$ and I guess we have a matrix with $1$'s on all values above the primary diagonal? But this isn't nilpotent degree 2 –  Mar 11 '15 at 04:45
  • The choices are two 2 by 2 Jordan blocks or a single 2 by 2 and a pair of 1 by 1's. So, either two ones in positions $1,2$ and $3,4,$ (they must be separated) or just one of those, everything else zero. – Will Jagy Mar 11 '15 at 04:53
  • Oh, by $1,2$ I mean row 1, column 2. The other is row 3, column 4. – Will Jagy Mar 11 '15 at 04:57

2 Answers2

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There are more forms. You will do well in considering Jordan blocks and in comparing the minimal polynomial to the characteristic polynomial.

One such example:

$$\begin{bmatrix}0&0&a&0\\0&0&0&0\\0&0&0&0\\0&b&0&0\end{bmatrix}=0_{4\times 4}$$

Nothing special
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It's not clear what the post means by "families". I assume it's about $4\times 4$ nilpotent matrices of degree $2$ up to similarity i.e., similarity class of matrices such that $A\neq \mathbf 0$ but $A^2=\mathbf 0$.

Read this answer (and also the question post) for context.

Up to similarity, there are two $4\times 4$ matrices with degree of nilpotence $2$ corresponding to the partitions $2+2$ and $2+1+1$.

$\displaystyle \begin{pmatrix} 0 & 1 & & \\ 0 & 0 & & \\ & & 0 & 1\\ & & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 1 & & \\ 0 & 0 & & \\ & & 0 & \\ & & & 0 \end{pmatrix}\tag*{}$

All matrices in the described family are similar to one of these.

Note that the blank entries are zeroes.

Nothing special
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