# How to show that the nth power of a $n \times n$ nilpotent matrix equals to zero $A^n=0$

$$A$$ is a $$n\times n$$ matrix such that $$A^m = 0$$ for some positive integer $$m$$. Show that $$A^n = 0$$.

My attempt:
For $$n > m$$, it's obvious since matrix multiplication is associative.

For $$n < m$$, $$A^n\times A^{m-n} = 0$$; not sure what to do next. Also I know that $$\det A = 0$$.

If $A^m=0$, then the minimal polynomial of $A$ divides $x^m$. The minimal polynomial has degree at most $n$ by Cayley-Hamilton.

Here is an alternative approach that doesn't rely on knowing anything about minimal polynomials, or Cayley-Hamilton. Consider $A$ as a linear transformation on $K^n$ if $K$ is your base field. Let $M_1$ be the range of $A$, i.e., $M_1=A(K^n)=\{Av:v\in K^n\}$. For $j>1$ let $M_{j}=A(M_{j-1})$ be the image of $M_{j-1}$ under $A$. In other words, $M_j=A^j(K^n)$, and we can include $M_0=K^n$ for convenience. Note that each $M_j$ is a subspace and $M_j\subseteq M_{j-1}$ for all $j$.

We know that $A^m=0$, so $M_m=\{0\}$. If for some $j$, $M_j=M_{j-1}$, then $M_{j+1}=AM_{j}=AM_{j-1}=M_j$, so $M_{j-1}=M_j=M_{j+1}=M_{j+2}=\cdots$. It follows that $M_j$ must be $\{0\}$ in such cases. This implies that all of the containments are proper until you get to $\{0\}$, so there is an $m_0$ such that $\{0\}=M_{m_0}\subsetneq M_{m_0-1}\subsetneq\cdots\subsetneq M_2\subsetneq M_1\subsetneq K^n$. Since there are $m_0$ proper containments of subspaces and $K^n$ is $n$ dimensional, this implies that $m_0\leq n$. Since $M_{m_0}=\{0\}$ and $M_{m_0}$ is the range of $A^{m_0}$, it follows that $A^{m_0}=0$, and finally $A^n=A^{m_0}A^{n-m_0}=0$.

From $A^m=0$ you learn that all eigenvalues of $A$ are zero. So the Jordan form of $A$ is strictly upper triangular, and it is an easy exercise to show that the $n^{\rm th}$ power of an upper triangular matrix with zero diagonal is zero.

• this matrix may not be diagonalizable. So to prove that its nth power is 0 i'm not quite sure how Feb 12, 2012 at 7:11
• @imintomath: Nobody suggested diagonalizing. Have you seen Jordan form? Feb 12, 2012 at 7:16
• o.o oh i haven't learn about this Jordan form yet, but now it's quite obvious that $A^n = 0$. Thank you all ~ Feb 12, 2012 at 7:29
• @martin..can you please elaborate why all the eigenvalues of A are zero? Although A is singular. Feb 4, 2019 at 15:53
• For any polynomial $p$, the eigenavlues of $p(A)$ are $p(\lambda)$ for $\lambda$ eigenvalue of $A$. Feb 4, 2019 at 16:46

Since this question is marked as duplicate from one that explicitly asks not to use the Cayley-Hamilton theorem I will not do that. Nor will I use Jordan normal forms (which require an algebraically closed field).

Let $$m\in\Bbb N$$ be minimal such that $$A^m=0$$; then by assumption there are (unless $$m=0$$, which implies $$n=0$$ and is boring) vectors$$~v$$ such that $$A^{m-1}\cdot v\neq0$$; one needs to prove $$m\leq n$$. There are various ways to proceed, I'll mention three.

(1) Putting $$v_i=A^{m-i}\cdot v$$ for $$0, one has $$v_i\in\ker(A^i)\setminus\ker(A^{i-1})$$, which shows that the chain of subspaces $$\{0\}=\ker(A^0)\subseteq\ker(A^1)\subseteq\ker(A^2)\subseteq\cdots$$ is strictly increasing until reaching $$\ker(A^m)=k^n$$, and considering dimensions one gets $$n\geq m$$.

(2) Those vectors $$v_1,\ldots,v_m$$ are linearly independent, for otherwise, if one considers the first $$k$$ for which $$v_1,\ldots,v_k$$ are linearly dependent, then $$k>1$$ (since $$v_1\neq0$$ by construction), but applying $$A$$ to a dependency relation one gets a shorter one, contradicting minimality. From the independence one gets $$m=\dim\langle v_1,\ldots,v_m\rangle\leq n$$.

(3) Putting $$V_i=\ker(A^i)$$ for brevity (any $$i\in\Bbb N$$), each $$V_{i+1}$$ is (by definition) the inverse image under the linear map defined by $$A$$ of $$V_i$$. Then that linear map induces a map $$V_{i+2}/V_{i+1}\to V_{i+1}/V_i$$ that is injective (by the mentioned inverse image property). It follows that the sequence of natural numbers $$(\dim(V_{i+1}/V_i))_{i\in\Bbb N}$$ is weakly decreasing, so it cannot attain the value $$0$$ for any $$i. Then immediately $$\dim(V_i)\geq i$$ for all $$i\leq m$$, and in particular $$n=\dim(V_m)\geq m$$. [The main interest of this approach is that it gives additional and quite useful information beyond what was asked for.]

Here is a matrix-theoretic proof. We use mathematical induction on $$n$$. The base case $$n=1$$ is trivial. In the inductive case, if $$A=0$$, there is nothing to prove. Suppose $$A\ne0$$. Therefore $$m\ge2$$.

Let $$r$$ be the rank of $$A$$. Then $$1\le r and $$A$$ admits a rank factorisation $$UV^T$$ where $$U$$ and $$V$$ are $$n\times r$$ matrices with full column rank. Now $$0=A^m=(UV^T)^m=U(V^TU)^{m-1}V^T$$. Since $$U$$ and $$V$$ have full column rank, $$U$$ has a left inverse and $$V^T$$ has a right inverse. It follows that $$(V^TU)^{m-1}=0$$. Thus $$V^TU$$ is an $$r\times r$$ nilpotent matrix. Yet, by induction assumption, we must have $$(V^TU)^r=0$$. It follows that $$A^{r+1}=U(V^TU)^rV^T=0$$. Since $$r+1\le n$$, we conclude that $$A^n=0$$.