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Let $V$ be a finite-dimensional $K$-vector space and let $T$ be a linear operator from $V$ to $V$.

I already proved that for every polynomial $p(x) \in K[x]$, $\ker p(T)$ and $Im p(T)$ are $T$-invariant. Is it true that for every $W$ proper $T$-invariant subspace, $W$ is the kernel of $p(T)$ for a polynomial $p$?

My guess is yes. If $p$ is the characteristic polynomial of the restriction of $T$ to $W$, then $p(T)v=0$ for every $v$ and then $W\subset \ker p(T)$. How can I finish the converse?

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Hint: What happens when $T$ is the identity operator $I$? What can $p(I)$ be? What can $\ker p(I)$ be? Which subspaces are $I$-invariant?

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  • $\begingroup$ Thanks a lot, I actually saw your first answer but it didn't occurred to me such obvious counter example $\endgroup$ – Jonas Gomes Jan 29 '15 at 16:19

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