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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.

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Thanks for point my mistake, I edit it just now ;) –  Mahan Feb 16 '12 at 22:19
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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. –  Carl Mummert Feb 16 '12 at 22:54
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1 Answer

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

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;) neat comment ! Is there any chance to solve it with eigenvalues and eigenvectors ? –  Mahan Feb 16 '12 at 22:32
If you are sure there are eigenvalues, for example if you work on $\mathbb C$, then and eigenspace for $T$ is $T$ stable. –  Davide Giraudo Feb 16 '12 at 22:37
@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. –  Arturo Magidin Feb 16 '12 at 22:49
Over $\mathbb{C}^n$, every linear transformation has a non-trivial, proper invariant subspace anyway, namely the span of a single arbitrary eigenvector. –  Carl Mummert Feb 16 '12 at 22:52
@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... –  fpqc Feb 16 '12 at 22:58
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