# Linear algebra: cyclic vector problem

Let $T:V \rightarrow V$ linear transformation over field $F$. Let $W$ be a non zero proper $T$-invariant subspace of $V$. Suppose that the characteristic polynomial $f_T$ of $T$ satisfies that $f_T(0)\neq 0$. Show that if $T$ has a cyclic vector, then the restricted $T|_W:W \rightarrow W$ has a cyclic vector.

I think I should use that $p_T=f_T$ but how can I connect this with the invariant space?

• The condition on the characteristic polynomial is redundant; it follows automatically from the fact that $T$ has a cyclic vector. – user29743 Jul 31 '12 at 18:58

The fact that $f_T(0)\ne 0$ means that $T$ is an isomorphism. Otherwise, there'd be a nonzero vector $x\in V$ with $T(x)=0$, implying $f_T(0)=0$.
Since $W$ is invariant, there exists a $T$-invariant complement $U$ of $W$ in $V$ (this might need some further argument, but it is true), i.e. $V=W\oplus U$. Since $T$ is an isomorphism and $T(W)=W$, we must have $T(U)=U$. Let $v=w\oplus u$ a cyclic vector for $T$ with $w\in W$ and $u\in U$. Then, we have $T^i(v)=T^i(w)\oplus T^i(u)$ with $w_i:=T^i(w)\in W$ and $T^i(u)\in U$ for all $i$. These vectors span $V$, so the projections $w_i$ for $0\le i< n$ must span $W$. Let $W_i$ denote the span of the vectors $w_0,\ldots,w_i$. If $T(w_i)\in W_i$, then we have $T^k(w_i)\in W_i$ for all $k\ge 1$. Since the $w_i$ span $W$, this can only happen if $W_i=W$. Hence, we have shown that $w$ is cyclic for $T|_W$.
• It's not quite true that $TU=U$ for every complement $U$ to $W$ in $V$, so the argument needs to be refined. – paul garrett Jul 31 '12 at 17:23
• Yea, that's true. I hope this edit makes it work - we don't need $U$ to be invariant if I am not mistaken. – Jesko Hüttenhain Jul 31 '12 at 17:42
• Hm, still the notion of "projection" (to $W$) is delicate, depending on a choice of complement. – paul garrett Jul 31 '12 at 17:52
• Yea, that was not such a good idea. There definitely exists a $T$-invariant complement of $W$ in $V$. One can see this using Jordan Normal Form, but I have a gut feeling that the OP does not want to assume that. I will give it some thought and try to find an elementary argument. – Jesko Hüttenhain Jul 31 '12 at 18:36
Perhaps it is possible to give a more elementary argument, but one usual argument is to use the structure theorem for finitely-generated modules over PIDs to say that $V$ is isomorphic to $\bigoplus_i F[x]/d_i$ where $x$ acts on $V$ by $T$, and $d_1|\ldots|d_m$ are monic polynomials, the "elementary divisors" of $T$. The assumption that there is a cyclic vector is that there is just one elementary divisor. From this viewpoint, it is clear that the property is inherited by any subspace.