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I want to find all nilpotent $2\times 2$ matrices.

All nilpotent $2 \times 2$ matrices are similar($A=P^{-1}JP$) to $J = \begin{bmatrix} 0&1\\0&0\end{bmatrix}$

But how do I find all of these matrices?

I do think that the only such cases are $J$ and $J^ T$

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  • $\begingroup$ Consider $2J$. That's nilpotent, too. Why not just write down an arbitrary matrix $P$, compute its inverse and the product $A$ you've written above, and set it equal to $J$ and solve? You get a system of 4 quadratics, it's true...but I'm guessing they won't be too bad. Have you tried that? Where did it lead? $\endgroup$ – John Hughes Mar 22 '15 at 11:47
  • $\begingroup$ @John I haven't, good idea.(Above I meant any scalar multiple of $J$ in response to the other statement). I will do that now. $\endgroup$ – user142198 Mar 22 '15 at 11:48
  • $\begingroup$ @lhf Could you clarify parametrization in this context please? I haven't seen the term used outside of calculus $\endgroup$ – user142198 Mar 22 '15 at 11:52
  • $\begingroup$ @rschwieb Indeed, good catch. $\endgroup$ – user142198 Mar 22 '15 at 12:03
  • $\begingroup$ @lhf I guess there is a typo in "all nilpotent matrices subject to ad−bc≠0" because the set is empty. Nilpotent matrices will certainly always have determinant zero... But yeah, the suggestion to simply compute $A^2$ for a square matrix certainly leads to a parametrization. $\endgroup$ – rschwieb Mar 22 '15 at 12:04
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My mistake... in my comment above, I should not have said to solve the equations.

Rather: if you pick an arbitrary $P$, and compute $P^{-1} J P$, you get a nilpotent matrix. Using

$$ P = \begin{bmatrix} a & b \\ c & d\end{bmatrix} $$ but writing $P J P^{-1}$, I got $$ PJP^{-1} = D \cdot \begin{bmatrix} -ac & a^2 \\ -c^2 & ac\end{bmatrix} $$ where $D$ is the determinant $(ad - bc)$.

Such matrices (for $a \ne 0$) have the general form

$$ A = \begin{bmatrix} -S & T \\ -\frac{S^2}{T} & S\end{bmatrix}. $$ and this is a "parameterization" (with parameters $S$ and $T$) of almost all possible nilpotent matrices. We also need to add in the $a = 0$ case, i.e. $$ A = \begin{bmatrix} 0 & 0 \\ c & 0\end{bmatrix}. $$

Note that in this parameterization, it's essential that $T \ne 0$. Because of this, you can say that up to scalar multiples, all nilpotent matrices have the form $$ A = \begin{bmatrix} -S & 1 \\ -S^2 & S\end{bmatrix}. $$ or $$ A = \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}. $$

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  • $\begingroup$ Oh indeed it works. Thank you $\endgroup$ – user142198 Mar 22 '15 at 12:01
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    $\begingroup$ But $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ is nilpotent and is not a multiple of any matrices of the form $\begin{pmatrix}-S&1\\-S^2&S\end{pmatrix}$ for any scalar $S$. $\endgroup$ – gniourf_gniourf Mar 22 '15 at 13:24
  • $\begingroup$ Correct. I ignored the $a = 0$ case. Sorry 'bout that. I've now fixed up the answer. $\endgroup$ – John Hughes Mar 22 '15 at 13:41
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Suppose that $A$ ($2\times 2$) is nilpotent. Then $\det(A)$ is $0$, implying that an eigenvalue of $A$ is real and $0$. Because $A$ is $2\times 2$, there is one other eigenvalue which must also be $0$. (If $A^n=0$, then the eigenvalues of $A^n$ are $0$ but these are just the eigenvalues of $A$ raised to the same power $n$.) At this point, you can infer that $\text{tr}(A)=0$ and $A$ itself must necessarily take the form $$ A=\begin{pmatrix}x & a \\ b & -x\end{pmatrix},\quad x\in\mathbb{R},\quad ab=-x^2. \tag{$*$} $$ It turns out that ($*$) is also sufficient for nilpotency: $$ \begin{pmatrix}x & a \\ b & -x\end{pmatrix}\cdot \begin{pmatrix}x & a \\ b & -x\end{pmatrix}=\begin{pmatrix}x^2+ab & xa-ax \\ bx-xb & ab+x^2\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}\cdot $$ Thus $A$ is nilpotent iff $A$ takes the form given in ($*$).

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    $\begingroup$ Indeed, characteristic equation for $2 \times 2$ matrices $A^2-tr(A)A+\det(A)I=A^2=0$ if $tr(A)=0$ and $\det(A)=0$ $\endgroup$ – Widawensen Mar 13 '18 at 13:27

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