# Cyclic subspace problem

Let $A$ be an $n\times n$ matrix with $\mathbb{C}$ entries and suppose the minimal polynomial of $A$ is $m(t)=t^n$. Show that there exists a $v\in V$ such that $\{v,Av,...,A^{n-1}v\}$ spans $V$.

My attempt: Since deg$(m)=n$, we must have that the minimal polynomial and characteristic polynomial $p$ are equal. Also, $\lambda=0$ is the only eigenvalue of $A$. In general, if $m$ has a factor $(t-\lambda_i)^{d_i}$, then $d$ is the size of the largest $\lambda_{i}$-Jordan block, and if $p$ has a factor $(t-\lambda_i)^{c_i}$ then $c_i$ is the number of times $\lambda_i$ appears in the Jordan normal form. All of this implies that the Jordan form for $A$ has $0$ in every entry except the superdiagonal, which consists of $1$s. If $\{j_1,...,j_n\}$ is the Jordan basis for this matrix, then $j_n$ is the desired vector. For example: if $n=4$, then we have $A\sim\begin{pmatrix} 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0 \end{pmatrix}$

thus $Aj_4=j_3\implies A^2j_4=Aj_3=j_2\implies A^3j_4=Aj_2=j_1$.

Does this look ok?

• Looks ok to me. This exercise is often given before Jordan canonical forms are covered. For extra credit: can you think of a solution that does not user Jordan forms? Hint: Let $v$ be a vector such that $A^{n-1}v\neq0$. – Jyrki Lahtonen Aug 1 '16 at 21:38
• I was actually going to ask if someone had a solution that didn't use Jordan normal form. I'll see if I can think of one – user124910 Aug 1 '16 at 21:48

Since $A^n = 0$ and $A^{n-1} \neq 0$, we can choose $v \in \mathbb{C}^n$ with $A^{n-1}v \neq 0$. Let us show that $(v,Av,\dots,A^{n-1}v)$ is linearly dependent and hence spanning $\mathbb{C}^n$. Assume that
$$a_0 v + a_1 Av + \dots + a_{n-1} A^{n-1}v = 0.$$
Applying $A^{n-1}$ to both sides of the equation, we get
$$a_0 A^{n-1}v + a_1 A^n v + \dots + a_{n-1} A^{2n - 2}v = a_0A^{n-1}v = 0.$$
Since $A^{n-1}v \neq 0$, we must have $a_0 = 0$. Repeating the argument with $A^{n-2}$ instead we see that $a_1 = 0$ and so on showing that $a_0 = \dots = a_{n-1} = 0$.