# How can I construct a nilpotent matrix with the property $A^2 \not= 0$ but $A^3=0$

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?

• You need bigger matrices for this. Try $3\times3$ upper triangular ones. – Jyrki Lahtonen Jan 12 at 10:19
• Like this: $\huge \begin{bmatrix} 0 &1 \\ 0&0\end{bmatrix}$ – Git Gud Jan 12 at 10:22
• @JyrkiLahtonen thanks! – Nullspace Jan 12 at 10:39

This is only possible in a space of dimension $$\ge k$$. In such a space take the matrix

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

• 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? – Nullspace Jan 12 at 10:26
• 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. – mathcounterexamples.net Jan 12 at 10:27
• 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. – Git Gud Jan 12 at 10:27
• I’m speaking of the vector space on which the matrices operate. Not the space of the matrices. – mathcounterexamples.net Jan 12 at 10:29
• That vector space here is, arguably, either $\mathbb R$ or $\mathbb C$, both of which have dimensions smaller than $3$. – Git Gud Jan 12 at 10:30

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 $$A^2=0$$.

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.