Nilpotent Matrix

Question

So I saw this problem: Is there an upper triangular matrix $A$ such that $A^n\neq 0$ but $A^{n+1}=0$? Prove or disprove.

I said no, and my reasoning was that the matrix must have a zero diagonal since $(A^k)_{ii}=a_{ii}^k$. Then the matrix must be srtictly upper diagonal, and when we multiply it the diagonal of zeros starts "moving up" and eventually the matrix is zero and it follows that it must be in at most $n$ steps.

Is there a neater way to do this? Like looking at characteristic polynomials or something? I was thinking that the minimal polynomial should be of the form $t^k$ and thus the characteristic polynomial ought to be $t^n$ as the minimal divides the characteristic, and by Cayley Hamilton we ought to have $A^n=0$, but I cannot see why the minimal polynomial must be $t^k$. If I was working over a complex (or alg closed field then yes since it could only have zero eigenvalues). Any thoughts?

Thanks,

Answer 1

There are a couple of ways of interpreting this question. The two most obvious ones (well, to me):

• Given $n\gt 0$, is there an upper triangular square matrix $A$ of some size (which may depend on $n$) such that $A^n\neq 0$ but $A^{n+1}=0$?

  The answer to this is "yes", and in fact you can take $A$ to be $(n+1)\times(n+1)$. One possibility is $$A = \left(\begin{array}{cccccc} 0 & 1 & 0 &\cdots & 0 & 0\\ 0 & 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 0 & 1\\ 0 & 0 & 0 & \cdots & 0 & 0 \end{array}\right).$$ If $\mathbf{e}_i$ is the vector with $1$ in the $i$th coordinate and $0$s elsewhere, then $A\mathbf{e}_1 = \mathbf{0}$, and $A\mathbf{e}_{k+1} = \mathbf{e}_k$ for $k=1,\ldots,n$. In particular, $A^n\mathbf{e}_{n+1} = \mathbf{e}_1\neq \mathbf{0}$, so $A^n\neq 0$; but $A^{n+1}(\mathbf{e}_i) = \mathbf{0}$ for all $i$, so $A^{n+1}=\mathbf{0}$. In fact, any strictly upper triangular $(n+1)\times(n+1)$ matrix will do.

• Given $n\gt 0$, is there an $n\times n$ matrix $A$ such that $A^n\neq 0$ but $A^{n+1} = 0$?

  The answer here is "no". You are right that all eigenvalues would have to be equal to $0$: a slightly more formal proof is that if $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigevalue of $A^k$ (check this! it's not hard); since the only eigenvalue of $0$ is $0$, if $A^{k}=0$ for some $k$, then every eigenvalue of $A$ is equal to $0$. Over the complex numbers (or the algebraic closure of your ground field), that means that the characteristic polynomial of $A$ must be $t^n$; which means that the characteristic polynomial over whatever field you are working on to begin with is also $t^n$, because all the roots in the algebraic closure already lie in the original field. This in turn means that, by the Cayley-Hamilton Theorem, we must have $A^n=0$. So we cannot have $A^n\neq 0$ and $A^{n+1}=0$: if $A^k=0$ for some $k$, then $A^n=0$.

  (By the way: by the definition of the minimal polynomial $m(t)$, if $p(t)$ is any polynomial such that $p(A)=0$, then $m(t)$ must divide $p(t)$; for if we divide $p(t)$ by $m(t)$, $p(t) = q(t)m(t)+r(t)$ with $r=0$ or $\deg(r)\lt\deg(m)$, then since $A$ centralizes the coefficients we get $0=p(A) = q(A)m(A)+r(A) = r(A)$; since $m(t)$ is the monic nonzero polynomial of smallest degree that evaluates to $0$ on $A$, we must have $r(t)=0$, so $m(t)|p(t)$). The Cayley Hamilton Theorem then implies directly that the minimal polynomial must divide the characteristic polynomial. Since the characteristic polynomial is $t^n$, the minimal polynomial must be $t^k$ for some $k$, $1\leq k\leq n$.

  In fact one can prove that every irreducible factor of the characteristic polynomial will divide the minimal polynomial, and the minimal polynomial divides the characteristic polynomial.)