# Minimal polynomial of restriction to invariant subspace divides minimal polynomial

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 $T$.

• 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. Commented Sep 20, 2013 at 0:53

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!!!

• But how do you know $m(x)$ has higher degree than $m_W(x)$ Commented Oct 27, 2015 at 16:15
• @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. Commented Oct 27, 2015 at 16:49

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$~\mu_S$.