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I want to prove the following statement:

Let linear operator $T:V\rightarrow V$ ($V$ is $n$-dimentional). Then there exists a vector $v$ such that $\left\{ v, Tv, ..., T^{n-1}v \right\}$ form a basis of $V$, iff $f_T=m_T$.

(Actually, the $\Rightarrow$ direction is quite easy. I need to prove the other direction.)

I've seen couple of answers to this problem in this site, all using the "Rational Canonial Form", which is something I wasn't tought in the course (but I am familiar with the Jordan Form).

Moreover, the only proof I've found on the internet is this, which seemed to be promising, until I got to this sentence, which unfortunately seems to be a mistake:

See that $V_i$ is a subspace of $V$, for all $1 \leq i \leq m$, and $V=\bigcup_{i=1}^m V_i$. Therefore $V=V_k$, for some $1 \leq k \leq m$.

I don't know if that's true for infinite vector spaces, but for finite ones, such as $\mathbb{F}_p^n$ it's certainly isn't.

Any step towards a proof would be appriciated (as well as an explanation for the suspicious claim stated above). Thank you!

Yiorgos S. Smyrlis
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35T41
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1 Answers1

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Suppose that the degrees of the characteristic polynomial $p(x)$ and minimal polynomial $m(x)$ of $T:V\to V$ coincide.

One can express $V$ as a direct sum of the generalised eigenspaces of $T$. If $\lambda_1,\ldots,\lambda_k$ are the eigenvalues of $T$, then $$ V=V(\lambda_1)\oplus\cdots\oplus V(\lambda_k), $$ where $V(\lambda_j)=\{v\in V: (T-\lambda_j)^\ell v=0\,\,\text{for some $\ell\in\mathbb N$}\}$. Each of these generalised eigenspaces is $T-$invariant. Set $T_j$ the restriction of $T$ to $V(\lambda_k)$, and let $p_j(x)$ and $m_j(x)$ the characteristic and minimal polynomial of $T_j$, respectively. In fact we have that $p(x)=p_1(x)\cdots p_k(x)$ and $$ m(x)=lcm\{m_1(x),\ldots, m_k(x)\}=m_1(x)\cdots m_k(x), $$ since the $m_j$'s are pairwise prime. And since, $m(x)$ and $p(x)$ have the same degree, then $m_j(x)=(x-\lambda_j)^{\mu_j}$, where $\mu_j$ is the multiplicity of $\lambda_j$ in $p(x)$.

Hence $(T_j-\lambda_j)^{\mu_j-1}\ne 0$ and $(T_j-\lambda_j)^{\mu_j}= 0$. Thus there exists a $v_j\in V(\lambda_j)$, such that $$ V(\lambda_j)\ni (T_j-\lambda_j)^{\mu_j-1}v_j\ne 0. $$ In fact $\{v_j,Tv_j,\ldots,T^{\mu_j-1}v_j\}$ is a basis of $V(\lambda_j)$.

Finally set $$ v=v_1+\cdots+v_k. $$ Then $\{v,Tv,\ldots,T^{n-1}v\}$ is a basis of $V$.

Indeed, if $$ 0=c_0v+c_1Tv+\cdots+c_{n-1}T^{n-1}v=p(T)v, $$ then $p(T)v_j=0$, for all $j=1,\ldots k$. Hence $(x-\lambda_j)^{\mu_j}$ divides $p(T)$, and hence the characteristic polynomial of $T$ divides $p$. Thus $p\equiv 0$, since $\deg p<n$.

Yiorgos S. Smyrlis
  • 86,581
  • Thank you very much for your answer. I would like to ask you about two clarifications: – 35T41 May 21 '17 at 06:55
  • You assume that the characteristic polynomial splits, right? But that's fine because we can always extend a field to its algebraic closure and it can't "destroy" our cyclic basis.
    • – 35T41 May 21 '17 at 06:55
  • Why is the sum of those cyclic vectors cyclic? Assume we have $v_1, Tv_1$ as a basis for one space and $v_2, Tv_2$ as basis for the second one. Then our list is $v_1+v_2, Tv_1+Tv_2, T^2v_1+T^2v_2, T^3v_1+T^3v_2$. Who said that it works?
    • – 35T41 May 21 '17 at 06:58
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    For the second clarification. Note that $T^kv_j\in V(\lambda_j)$, for all $k$. Assume that $$ p(T)v=c_0v+c_1Tv+\cdots+c_{n-1}T^{n-1}v=0.$$ Then $ p(T)v_j=0$ and hence $(x-\lambda_j)^{\mu_j}$ divides $p(T)$. – Yiorgos S. Smyrlis May 21 '17 at 10:03
  • @35T41 It would be challenging to prove this statement without the use of eigenvalues and generalised eigenspaces, as this requires algebraically closed fields. – Yiorgos S. Smyrlis May 21 '17 at 12:27
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    Why is that so challenging? Just replace the irreducible factors $X-\lambda$ by more general monic irreducible polynomials. – Marc van Leeuwen May 22 '17 at 14:09