$A$ 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 $\det A = 0$.


4 Answers 4


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.

  • $\begingroup$ this matrix may not be diagonalizable. So to prove that its nth power is 0 i'm not quite sure how $\endgroup$ Feb 12, 2012 at 7:11
  • 3
    $\begingroup$ @imintomath: Nobody suggested diagonalizing. Have you seen Jordan form? $\endgroup$ Feb 12, 2012 at 7:16
  • 2
    $\begingroup$ 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 ~ $\endgroup$ Feb 12, 2012 at 7:29
  • $\begingroup$ @martin..can you please elaborate why all the eigenvalues of A are zero? Although A is singular. $\endgroup$ Feb 4, 2019 at 15:53
  • $\begingroup$ For any polynomial $p$, the eigenavlues of $p(A)$ are $p(\lambda)$ for $\lambda$ eigenvalue of $A$. $\endgroup$ 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 $0<i\leq m$, 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 $i<m$. 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 $1\le r<n$ 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 $A^n=0$.


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