Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Full version of the problem is following:

Let T be a linear transformation on a finite dimensional vector space $V$ over a field $\mathbb{F}$. If the minimal polynomial $p_t$ of T is irreducible, then every T invariant subspace $W$ has a T-invariant complement $W'$

I used Cyclic decomposition Theorem which states that

"The finite dimensional vector space V can be expressed a s a decomposition of T-cyclic subspaces $Z(\alpha_1;T)\oplus Z(\alpha_2;T)\oplus...\oplus Z(\alpha_k;T)$ and their annihilators $p_1,...p_k$ have properties; (1) $p_k|...|p_1$, (2)$p_T=p_1, f_T=p_1\cdot p_2\cdot ...\cdot p_k$ where $f_T$ is the characteristic polynomial of T."

Can I say this? Since $p_T$ is irreducible, there is a cyclic vector $\alpha$ such that $V=Z(\alpha;T)$ and V=W and $W'=\{0\}$. Therefore $W'$ for each W is T-invariant.

Is my way correct?

Thank you in advance.

share|cite|improve this question
Certainly the $W=V$ step doesn't make any sense. If, say, $W$ had 5 dimensions, then it would have many $T$ invariant subspaces of dimension 1,2,3,4... are those all $V$ as well? Please also consider explaining some of your notation. – rschwieb Aug 12 '12 at 0:47
up vote 4 down vote accepted

Let $\mathbb{F}[X]$ be the polynomial ring with one variable. $V$ can be regarded as an $\mathbb{F}[X]$-module by defining $Xv = T(v)$ for every $v \in V$. $\mathbb{F}[X]$-submodules of $V$ are none other than $T$-invariant subspaces of $V$. Let $K = \mathbb{F}[X]/(p_t)$. Since $p_t$ is irreducible, $K$ is a field. Since $p_t V = 0$, $V$ can be regarded as a $K$-module. Let $W$ be a $T$-invariant subspace of $V$. $W$ can be regarded as a $\mathbb{F}[X]$-submodule of $V$. Since $p_t W = 0$, $W$ can be regarded as a $K$-submodule of $V$. Hence there exists a $K$-submodule $W'$ such that $V = W \oplus W'$. Since $W'$ is a $\mathbb{F}[X]$-submodule, it is $T$-invariant. This completes the proof.

share|cite|improve this answer

The answer by Makoto Kato really captures the essence of the answer best, but you can also do this without using a field extension. The condition that the minimal polynomial $p_T$ be irreducible is a rather strong condition: it means that for every nonzero vector$~v$ the minimal degree monic polynomial$~P$ such that $P[T](v)=0$ is equal to$~p_T$ (in general monic divisors of$~p_T$ are candidates, but here the only such proper divisor is$~1$, but it only annihilates the zero vector). In particular $v,T(v),\ldots,T^{d-1}(v)$ are always linearly independent, where $d=\deg(p_T)$. Consequently, the only $T$-invariant subspaces of the span$~S$ of these $d~$vectors are the zero space and$~S$ itself: this is because for every nonzero vector$~v'$, its repeated images $v',T(v'),\ldots,T^{d-1}(v')$ are linearly independent and therefore necessarily span$~S$.

Now a fundamental property is that in $V$, and by the same argument in any $T$-stable subspace of it (because the restriction of$~T$ to it has the same minimal polynomial, unless it has dimension$~0$), one can find a set of vectors $v_1,\ldots,v_k$ such that the $dk$ vectors $v_1,T(v_1),\ldots,T^{d-1}(v_1)$, $v_2,\ldots,T^{d-1}(v_2)$, $\ldots,v_k,T(v_k),\ldots,T^{d-1}(v_k)$ form a basis. In fact they can be chosen fairly freely, with the choice of $v_i$ being restricted only by being outside the span $S_{<i}$ the vectors before it in the list, as in the incomplete basis theorem. (In fact this is the incomplete basis theorem with respect to the $K$-vector space structure in the other answer I referred to.) To see this, it suffices to observe for the span $S_i$ of the repeated images $v_i,T(v_i),\ldots,T^{d-1}(v_i)$ that $S_{<i}\cap S_i=\{0\}$ (it is a $T$-stable subspace of $S_i$ that does not contain $v_i$) and one has a direct sum $S_{\leq i}=S_{<i}\oplus S_i$. The process ends when $S_{\leq i}=V$ in which case one sets $k=i$; note that we obtain that $\dim(V)=kd$ is necessarily a multiple of$~d=\deg(p_T)$.

Now for the result of the question: choose such a set of vectors $v_1,\ldots,v_k$ for the $T$-stable subspace$~W$, and complete to such a set of vectors $v_1,\ldots,v_l$ for the whole space; then it is easy to see that the vectors $v_{k+1},T(v),\ldots,T^{d-1}(v_{k+1})$, $\ldots,v_l,T(v_l),\ldots,T^{d-1}(v_l)$ are the basis of a $T$-stable complement of$~W$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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