Let $V$ be an n-dimensional vector space over $\mathbb{R}$ and $T:V \rightarrow V$ be linear. Call a vector $v \in V$ cyclic if $V$ is spanned by $\{v, \ Tv, \ T^2v,..\}$.


Show that $v$ is cyclic for $T$ then there is some $k \geq 1$ such that

$$B=\{v, \ Tv,..T^{k-1}v \}$$ is linearly independent in $V$ and that $T^kv$ lies in the span of $B$.


If we suppose that no such $k$ exists, then every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent, or it is linearly dependent with $T^lv$ not lying in it span.

There are in effect then two cases to consider.

Suppose every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent. Then $T^{l-1}$ can be expressed as a linear combination of $v,...,T^{l-2}v$. Now, $\{v, \ Tv,..T^{l-2}v \}$ is also linearly dependent by assumption and so we can express $T^{l-2}$ as a linear combination of $v,...,T^{l-3}v$ etc... Continuing in this way, we show that $T^{l-1}v= \lambda v$ for some $\lambda \in \mathbb{R}$. Thus $$B=\{ \mu v \ | \ \mu \ \text{is an eigenvalue of} \ T^mv \ \text{for some} \ m \}$$ This is 'at best' countably infinite, and therefore could not span $V$ even if it were 1-dimensional, wince our base field is $\mathbb{R}$, which is uncountable.

Suppose then that $B$ is linearly independent for some $k$ but that $T^kv$ does not lie in its span. I can't see how to progress this argument.

Note: This question is for second year Undergraduates, and I think there must be some more simple approach; even the first part surprises me if we need to argue using the uncountability of $\mathbb{R}$, so any alternative approaches are also appreciated.

  • $\begingroup$ You are well on your way. The simple fact you are thinking of is the Cayley-Hamilton theorem: every operator on a finite-dimensional vector space satisfies its own characteristic polynomial. In other words, there is a polynomial such that $(T^k + T^{k-1} + \dots + T + I)v = 0$ for all $v \in V$. $\endgroup$ – snar May 26 '13 at 17:41
  • $\begingroup$ You seem to be wandering in strange directions. If $v$ is nonzero then certainly $\{v\}$ is linearly independent, so you woorries that all sets $\{v,Tv,\ldots,T^kv\}$ would be dependent cannot possibly be realised. $\endgroup$ – Marc van Leeuwen May 26 '13 at 17:50
  • 1
    $\begingroup$ @snarski: Invoking the Cayley-Hamilton Theorem is, at best, making things about an order of magnitude more complicated than they need to be. Also your "in other words" is not the Cayley-Hamilton Theorem...e.g. because the Cayley-Hamilton Theorem, though of questionable relevance here, is a true mathematical statement, and yours is not: take $T$ to be the identity and $V = \mathbb{R}$, for instance. $\endgroup$ – Pete L. Clark May 26 '13 at 17:58

I had some trouble following your argument. You write

Suppose every subset of the form $\{v,Tv,\ldots,T^{\ell-1} v \}$ is linearly dependent.

If we are allowed to take $\ell = 1$ here then in particular you are assuming that $\{v \}$ is linearly dependent, and thus $v = 0$ so $V = 0$. Otherwise $\{v,Tv\}$ is linearly dependent so either $v = 0$ or $Tv = \alpha v$ for some scalar $\alpha$. If this is the case then $T^{\ell} v = \alpha^{\ell} v$ for all $\ell$, and the space $V$ must be one-dimensional, spanned by $v$. Nothing like the uncountability of the scalar field is needed for this.

But why have you started the argument this way? As the above analysis shows, you've done only a trivial case and an almost trivial case, so you're certainly less than halfway to a complete proof.

I suggest rather the following: let $\ell$ be the least positive integer such that $\{v,Tv,\ldots,T^{\ell} v \}$ is linearly dependent. Such an $\ell$ exists because $V$ is assumed to be finite-dimensional -- it need not exist otherwise. (We've handled the $\ell = 1$ case above.) This implies that $T^{\ell} v$ lies in the span of $1,Tv,\ldots,T^{\ell-1} v$, answering your question.

Note also that for all $L \geq \ell$, $T^{L} v$ lies in the span of $1,Tv,\ldots,T^{\ell-1} v$, so that $V$ itself is spanned by $1,Tv,\ldots,T^{\ell-1} v$. (At first glance I thought that this was your question. It turns out that it isn't, but it is the natural next thing to say, so you will surely be encountering it soon.)

The conclusion holds over any field of scalars; the real numbers have nothing to do with it.


For the question mentioned the fact that $v$ is a cyclic vector is irrelevant, it just needs to be nonzero (and that only because you insisted that $k\geq1$). Since $V$ is finite dimensional, the infinite sequence $v,Tv,T^2v,\ldots$ cannot remain linearly independent forever; taking $k$ minimal so that $v,Tv,\ldots,T^kv$ is linearly dependent, the conclusion of you question will be satisfied. Moreover the subspace $W$ spanned by $B$ will be stable under $T$ (that is, $T(W)\subseteq W$) so that it makes sense to restrict $T$ to a map $T|_W: W\to W$, and (and this is where the notion of cyclic vector comes in) $v$ is a cyclic vector in the subspace $W$ for the linear operator$~T|_W$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.