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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
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 :) – user21436 Feb 16 '12 at 22:21
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

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... – user38268 Feb 16 '12 at 22:58

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