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I need revision of the following problem:

"Let $T$ be a linear operator on a finite-dimensional space $V$. Suppose $p(x) = f(x) = p_1(x)^r$ where $p(x)$ is the minimal polynomial, $f(x)$ the characteristic polynomial and $p_1(x)$ is an irreducible polynomial. Show that there is no non trivial T-invariant subspace with complementary T-invariant subspace."

Now, this is my attempt:
Given a T-invariant subspace, the minimal polynomial for this subspace divides the minimal polynomial $p(x)$. So, consider the polynomial $p_1(x)^k$ with $2 \le k \lt r$ and define $W = \ker(p_1(T)^{k-1})$. $W$ is clearly T-invariant.
Since $p_1^k$ divides $p_1^r$ there exists a polynomial $h$ such that $$p(x) = p_1(x)^kh(x)$$ So we get that $p_1^{k-1}h(T)$ is not the zero operator, so for a $\beta \in V$, $\beta \ne 0$ we have $p_1^{k-1}h(T)\beta \ne 0$. It follows that $$p_1(p_1^{k-1}h(T))\beta = p_1^kh(T)\beta = p_1(T)^{k-1}(p_1h(T)\beta) = 0 $$ and therefore $p_1h(T)\beta \in W$. But there is no $\gamma \in W$ such that $p_1h(T)\beta = p_1h(T)\gamma$, because if there is such $\gamma$ then $$ p_1^{k-2}p_1h(T)\gamma = p_1^{k-2}p_1h(T)\beta \\ p_1^{k-1}h(T)\gamma = p_1^{k-1}h(T)\beta \\ \underbrace{hp_1^{k-1}(T)\gamma = 0 }_{\text{since $\gamma \in W$}} = p_1^{k-1}h(T)\beta$$ But $p_1^{k-1}h(T)\beta \ne 0$, a contradiction. So, for the different choices of $k$, $W$ cannot have a complementary T-invariant subspace.

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  • $\begingroup$ I think the claim is false. As a counterexample, have a look at$$T = \begin{pmatrix}1 & 0\\0 & 2\end{pmatrix}.$$ Then $f(x) = (1-x)(2-x) = q(x)$ and this polynomial is irreducible. However, $V = span\,\{e_1\}\oplus span\,\{e_2\}$ and both subspaces are invariant with respect to $T$. $\endgroup$
    – amsmath
    Oct 10, 2017 at 21:39
  • $\begingroup$ @amsmath ohh I believe the minimal polynomial has to be a power of an irreducible polynomial in this case $\endgroup$
    – mate89
    Oct 10, 2017 at 23:38
  • $\begingroup$ Yes, sorry. I forgot what irreducible means for a moment... $\endgroup$
    – amsmath
    Oct 11, 2017 at 1:39

1 Answer 1

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Let $W$ a proper invariant subspace. The minimal polynomial of $T_{|W}$ divides the minimal polynomial of $T$, so must be of form $p_1^s$ for some $s< r$. If $W$ had an invariant complement, we would conclude $p_1^{u}$ annihilates $T$, for some $u<r$.

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  • $\begingroup$ Why $p_1^u$ annihilates T and not only the restriction of T under that complement? Can't see it .. $\endgroup$
    – mate89
    Oct 10, 2017 at 23:59
  • $\begingroup$ @mate89: $p_1^{u}$ annihilates both restrictions, hence also their direct sum. $\endgroup$
    – orangeskid
    Oct 11, 2017 at 0:25
  • $\begingroup$ @amsmath: the minimal polynomial of $T_{|W}$ has degree $< \dim V$ .Also it has to divide the minimal polynomial of $T$, $p^r$, of degree $\dim V$. So it must be a lesser power. $\endgroup$
    – orangeskid
    Oct 11, 2017 at 1:38
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    $\begingroup$ This was actually clear. My problem was that you did not describe the key point very clearly. We have minimal polynomials $p_1^s$ and $p_1^t$ for the two restrictions, where $s,t < r$. Hence $p_1^m$ annihilates $T$, where $m := \max\{s,t\} < r$. Contradiction! $\endgroup$
    – amsmath
    Oct 11, 2017 at 1:47
  • $\begingroup$ @amsmath: Yes, that makes it transparent! Good call. $\endgroup$
    – orangeskid
    Oct 11, 2017 at 1:52

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