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 $T$.

Solutions Collecting From Web of "Minimal polynomial of restriction to invariant subspace divides minimal polynomial"

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 algorirhm 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 $0 \le \deg r(x) < \deg m_W(x)$, whence $r(x) = m(x) – m_W(x)q(x)$. Then $r(T_{\vert W}) = m(T_{\vert W}) – m_W(T_{\vert W})q(T_{\vert W}) = 0$. 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

Cheers, and as always,

Fiat Lux!!!

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$~\mu_S$.