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I'm trying to prove this:

$T : V \to V$ linear transformation. $W$ subspace of $V$. If $W$ is $T$-invariant then the minimal polynomial for the restriction operator $T|_W$ divides the minimal polynomial for $T$.

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  • $\begingroup$ let m(x) be the minimal polynomial for T. I tried using the fact that m(T)=0 so the restriction of m(T) to W is 0. then we have that m(T)|W = m(T|W) but then I get stuck. I also tried using the fact that the minimal polinomial divides the characteristic polinomial and that the characteristic polinomial of T restricted to W divides the characteristc polinomial of T. $\endgroup$
    – Nikki
    Commented Sep 20, 2013 at 0:53

2 Answers 2

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I think this will work:

Suppose $V$ is a vector space over the field $\Bbb F$, with $\dim V = N < \infty$. Then the minimal polynomial of $T$ is the monic polynomial $m(x) \in \Bbb F[x]$ of least degree such that $m(T) = 0$ identically on $V$. As such, we have $m(T) = 0$ on $W$ as well, whence $m(T_{\vert W}) = 0$. Now let $m_W(x) \in \Bbb F[x]$ be the minimal polynomial of the restriction $T_{\vert W}$ of $T$ to $W$. Since $\Bbb F$ is a field, the usual division algorithm for polynomials holds in $\Bbb F[x]$. Thus we may write $m(x) = m_W(x)q(x) + r(x)$ for some unique $q(x), r(x) \in \Bbb F[x]$ with either $r(x) = 0$ or $0 \le \deg r(x) < \deg m_W(x)$, whence $r(x) = m(x) - m_W(x)q(x)$. Then $r(T_{\vert W}) = m(T_{\vert W}) - m_W(T_{\vert W})q(T_{\vert W}) = 0$. Now in the event $r(x) \ne 0$, let the leading coefficient of $r(x)$ be $\beta \in \Bbb F$. Set $r'(x) = \beta^{-1} r(x)$. Then $r'(x)$ is monic, $\deg r'(x) = \deg r(x)$, and furthermore $r'(T_{\vert W}) = \beta^{-1} r(T_{\vert W}) = 0$. But since $\deg r'(x) < \deg m_W(x)$, this contradicts the minimality of $m_W(x)$ unless $r'(x) = \beta^{-1}r(x) = 0$. Thus $r(x) = 0$ and hence $m_W(x) \mid m(x)$. QED

Note Added in Edit, Sunday 29 August 2021 10:16 PM PST: Though the above proof seems pretty straightforward to me, by and large, I feel the assertion made in the third sentence, that

"As such, we have $m(T) = 0$ on $W$ as well, whence $m(T_{\vert W}) = 0$,"

may be further clarified. We observe that the given $T$-invarinace of

$W \subset V \tag{1}$

allows us to write

$Tw \in W \tag 2$

for any

$w \in W; \tag 3$

also, for any such $w$, by definition we have

$T_{\vert W} w = Tw, \tag 4$

and thus by virtue of (2) we may write

$T_{\vert W}Tw = TTw = T^2w, \tag{4.5}$

and hence, in accord with (2)-(4.5), we further have

$T^2_{\vert W} w = T_{\vert W} T_{\vert W} w = T_{\vert W} Tw = TTw = T^2w; \tag 5$

indeed, at this point we may allow a simple induction to take over, assuming

$T^k_{\vert W} w = T^k w \tag{6}$

for some $k \in \Bbb N$ and any $w \in W$; then since $T^k_{\vert W} w \in W$ we may write

$T^{k + 1}_{\vert W} w = TT^k_{\vert W} w = TT^k w = T^{k + 1}w; \tag 7$

from this it is easy to see that for any

$p(x) \in \Bbb F[x] \tag 8$

$p(T_{\vert W})w = p(T)w \tag 9$

provided $w \in W$; and from this we readily conclude that

$m(T)w = m(T_{\vert W})w = 0 \tag{10}$

as well. End of Note.

Cheers, and as always,

Fiat Lux!!!

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    $\begingroup$ But how do you know $ m(x)$ has higher degree than $m_W(x)$ $\endgroup$ Commented Oct 27, 2015 at 16:15
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    $\begingroup$ @Clair Symmetry: by definition, $m_W(x)$ is the monic polynomial of least degree such that $m_W(T_W) = 0$; since $m(T_W) = 0$ as well, it follows that $\deg m(x) \ge \deg m_W(x)$, and this is sufficient. And in fact, the proof I gave doesn't need this fact. $\endgroup$ Commented Oct 27, 2015 at 16:49
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For any linear operator $S$ of a space $W$, the set of polynomials $P$ such that $P[S]=0$ (on $W$) is the set (ideal in $K[X]$) of multiples of the minimal polynomial $\mu_S$ of $S$. Apply this in the question to the restriction $S=T|_W$, with $P=\mu_T$, the minimal polynomial of$~T$ on all of$~V$ (since $P[T]=0$ on all of$~V$, its restriction $P[S]$ is certainly $0$ on$~W$). This gives that $P=\mu_T$ is a multiple of$~\mu_S$.

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