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An example of a matrix $A$ that has the property $A^2=0$ would be $$A= \begin{pmatrix} 0 &1 \\ 0&0\end{pmatrix}$$

However, I can't seem to figure out a "formula" to construct a matrix that has the property $A^3=0$ but $A^2 \not = 0$. Or in general, a formula for a matrix that has the property $A^k=0$ with $A^{k-1} \not=0$. Does such a formula even exist?

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    $\begingroup$ You need bigger matrices for this. Try $3\times3$ upper triangular ones. $\endgroup$ – Jyrki Lahtonen Jan 12 at 10:19
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    $\begingroup$ Like this: $\huge \begin{bmatrix} 0 &1 \\ 0&0\end{bmatrix}$ $\endgroup$ – Git Gud Jan 12 at 10:22
  • $\begingroup$ @JyrkiLahtonen thanks! $\endgroup$ – Nullspace Jan 12 at 10:39
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This is only possible in a space of dimension $\ge k$. In such a space take the matrix

$$N_k=\begin{pmatrix} 0&1&0& \dots&0\\ 0&0&1& \dots&0\\ 0&\vdots&\ddots& \ddots&0\\ 0&0&\dots& 0&1\\ 0&0&0& \dots&0\\ \end{pmatrix}$$

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    $\begingroup$ So If I want to construct a nilpotent matrix that has the property $A^5=0$ but $A^4 \not=0$ I would have to start with a $5 \times 5$ upper triangular matrix? $\endgroup$ – Nullspace Jan 12 at 10:26
  • $\begingroup$ Yes, you need in that case to have a space of dimension $5$ at least. And the matrice I provided is an example of a possible choice. $\endgroup$ – mathcounterexamples.net Jan 12 at 10:27
  • $\begingroup$ I don't think your formulation is correct, "space of dimension greather than or equal to $k$". What space is this? The vector space of $k\times k$ matrices has dimension $k^2$, not $k$. Edit: Also, naming that matrix $I_k$ is a bad idea. $\endgroup$ – Git Gud Jan 12 at 10:27
  • $\begingroup$ I’m speaking of the vector space on which the matrices operate. Not the space of the matrices. $\endgroup$ – mathcounterexamples.net Jan 12 at 10:29
  • $\begingroup$ That vector space here is, arguably, either $\mathbb R$ or $\mathbb C$, both of which have dimensions smaller than $3$. $\endgroup$ – Git Gud Jan 12 at 10:30
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Not that you asked, but it's easy to see that this is impossible for a $2\times 2$ matrix. Say $A$ is $2\times2$ and $A^3=0$. If $p$ is the minimal polynomial of $A$ then $\deg(p)\le 2$ (by Cayley-Hamilton) and $p(t)|t^3$; hence $p(t)|t^2$, so $A^2=0$.

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You already got answer but I am just adding one thing which will help to understand the process.

For n=2, This linear transformation T(x,y)=(y,0) will give A nonzero but $A^{2}=0$.(perform linear transformation twice and you will see it and each LT is associated with matrix ,you will get matrix with that property.

Now for n=3, T(x,y,z)=(y,z,0) will satisfy given conditions.(perform LT thrice).Find it's matrix.

You can get pattern.

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