3
$\begingroup$

Linear transformation $T\colon V \to V$ has the property that there is no non-trivial subspace $W$ for which $T(W) \subseteq W$ . Prove that for every polynomial $P$ , $P(T)$ is either invertible or zero.

  • 0
    Thanks for point my mistake, I edit it just now ;)2012-02-16
  • 2
    And, you may consider accepting answers to your previous queries by clicking on the tick mark beside your favorite answer. Note that this is the only way to thank people who care for you on this site :)2012-02-16
  • 0
    The entire space is itself a non-trivial subspace $W$ for which $T(W) \subseteq W$. I'm certain that non-trivial *proper* subspaces are what was intended.2012-02-16

1 Answers 1

6

Hint: show that $\ker P(T)$ is a linear invariant subspace of $V$ using the fact that $TP(T)=P(T)T$.

  • 0
    ;) neat comment ! Is there any chance to solve it with eigenvalues and eigenvectors ?2012-02-16
  • 0
    If you are sure there are eigenvalues, for example if you work on $\mathbb C$, then and eigenspace for $T$ is $T$ stable.2012-02-16
  • 1
    @Mahan: If there is an eigenvalue $\lambda$, then the eigenspace is stable and hence must be all of $V$; therefore, $T$ is a scalar multiple of the identity, and $P(T)$ is the scalar matrix $p(\lambda)I$, hence either invertible or zero. But it's possible for $T$ to have no eigenvalues (e.g., a rotation by $90^{\circ}$ on the real plane), so you cannot solve it that way in general.2012-02-16
  • 0
    Over $\mathbb{C}^n$, every linear transformation has a non-trivial, proper invariant subspace anyway, namely the span of a single arbitrary eigenvector.2012-02-16
  • 0
    @ArturoMagidin Sounds like there is a connection with Schur's Lemma here. But then again that is only when $V$ is a complex vector space...2012-02-16
  • 0
    Well, it will only be proper if $n > 1$, of course.2012-02-16