Hint: 1) Show that there is a vector $e$ such that $(e,Ae,A^2e,\dots,A^{n-1}e)$ is a basis (this is another standard exercice).
Write $Be = \sum_{i=0}^{n-1} a_i A^i e$.
2) Show that $B$ and $\sum_{i=0}^{n-1} a_i A^i$ agree on the basis.
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!!!
Best Answer
Demo
Given an operator $\mathcal T$, the cyclic space generated by $v\in V$ is $$ C_v=\mathrm {span}\{v, \mathcal Tv, \dots, \mathcal T^k v, \dots\}. $$ Assume $V$ is finite dimensional, then there is some $p\in \mathbb N^*$ s.t. $(\mathcal T^j v)_0^{p-1}$ is a linearly independent set, but $(\mathcal T^j v)_0^{p}$ is not. So there exists a set of scalars $(a_j)_0^{p-1}$ s.t. $$ \mathcal T^p v = \sum_0^{p-1} a_j \mathcal T^j v. $$ Hence the polynomial $g(x ) = x^p - \sum_0^{p-1}a_j x^j$ satisfies that $g(\mathcal T)v = 0$, thus $g(x)$ annihilates $\mathcal T|_{C_v}$. Let $m(x)$ be the desired minimal polynomial, then since $g(\mathcal T)v =0$, $m|g$. On the other hand, if some monic polynomial $r(x) $ with degree $q <p$ satisfying $r(\mathcal T)v =0$, then $(\mathcal T^j v)_0^q$ is linear dependent, contradicting the definition of $p$. So $\deg m \geqslant \deg g$ where $m$ is an annihilating polynomial for $v$. Since $m, g$ are monic by definition, $m = g$, hence $\deg m = \deg g$. By the definition of $p$, we could conclude that $\dim(C_v)=p$, hence the desired conclusion.
UPDATE
Suppose $g(x)$ annihilates $v$, and $m(x)$ is the minimal one. Then perform the division with remainder, we could get polynomials $q(x), r(x)$ s.t. $g =mq + r$ where $\deg r < \deg m$. If $r \neq 0$, then since $m$ is the minimal one, $m$ annihilates $v$. Then $$ 0=g(\mathcal T)v = m(\mathcal T)q(\mathcal T)v + r(\mathcal T) = 0 + r(\mathcal T)v. $$ If $r \neq 0$, then $r$ is a polynomial annihilating $v$ with $\deg r <\deg m$, contradicting the minimality of $m$. Hence $g=mq $, equivalently $m | g$.
TEXTS I HAVE READ
A second course for LA. I think the choice of proofs are special.
A second course with theory developed without introducing determinants. It is pretty geometric.
A classic with deeply developed theories. Should be revisited after the 1st or 2nd exposure to the subject.
The crucial part is to use the polynomials to study the structure of linear spaces, which is covered by these books.