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Prove that for $n\times n$ matrix $A$ : $A^{n+1} = 0 \implies A^n = 0$

Is this statement true? How do you prove it?

Cave Johnson
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Jack Black
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  • Where do the entries of the matrix live? If you're not working over a field, the statement need not be true. You might find the theory of the characteristic/minimal polynomial of the matrix to be helpful. – Stahl Nov 21 '16 at 04:41
  • @астонвіллаолофмэллбэрг Why do you say the statement is definitely not true? It's certainly true over the reals, which is the domain I would assume where it isn't specified. – Erick Wong Nov 21 '16 at 18:46
  • @ErickWong I think I got confused. I did not realize that $n$ was fixed, I thought that it was arbitrary. I have deleted the comment. – Sarvesh Ravichandran Iyer Nov 24 '16 at 03:17

2 Answers2

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Assume not, there exists a non-zero eigenvalue $\lambda$, such that

$A^n*v=\lambda*v$

multiply $A$ on both sides, we have

$\lambda*Av=0$

Since $\lambda$ is not $0$, $Av=0$

hence $A^n*v=0=\lambda*v$, contradiction!

  • Underrated answer. – Decaf-Math Nov 21 '16 at 04:56
  • For clarification to the OP and anyone else coming here in the future--a matrix is singular if and only if $0$ is an eigenvalue. This is where the contradiction here happens--if you assume that $0$ is not an eigenvalue, then the steps shown in this answer imply $0$ is an eigenvalue which goes against what we supposed, hence you get the desired result. – Decaf-Math Nov 21 '16 at 04:59
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    @pyrazolam I'm sorry I don't quite follow your explanation. A matrix can be non-zero even though all of its eigenvalues are equal to 0. How does this show $A^n$ is the zero matrix? It doesn't appear that $n$ is used in this argument, yet the truth of the statement depends critically on $n$ (it is false for any exponent smaller than $n$). – Erick Wong Nov 21 '16 at 05:31
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    A concrete example of a nonzero matrix with all its eigenvalues equal to $0$ is the matrix which is $0$ except for its superdiagonal which is all $1$. – Patrick Stevens Nov 21 '16 at 09:07
  • Besides the problem that @Patrick Stevens pointed out, how would the last "hence" work? – Zhanxiong Dec 28 '22 at 00:52
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Use the fact that the sequence of the kernels is increasing and concave. More formally:

$$\forall i \in [|0,n-1|] \quad \ker(A^i) \subseteq \ker(A^{i+1})$$

and

$$\forall i \in [|0,n-2|] \quad \dim(\ker(A^{i+1})) - \dim(\ker(A^i)) \geq \dim(\ker(A^{i+2})) - \dim(\ker(A^{i+1}))$$

To be clearer: the sequence of the kernels is increasing (with respect to the set inclusion order), but if at the step $i \rightarrow i+1$ the kernel has gained $k$ dimensions, then it cannot gain more than $k$ dimensions at the following steps.

Now, $A^{n+1} = 0$ is equivalent to $\ker(A^{n+1}) = \mathbb{R}^n$, what can you conclude?

Edouard Berthe
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