The Minimal Polynomial

Part 1 (Existence). The vector space $\mathrm{End}(V)$ of linear operators on $V$ has dimension $n^2$ over $\mathbb{F}$ (it is isomorphic to the space of $n \times n$ matrices once a basis of $V$ is fixed). Consider the $n^2+1$ operators

These are $n^2+1$ vectors in an $n^2$-dimensional space, so they are linearly dependent: there exist scalars $a_0,\dots,a_{n^2} \in \mathbb{F}$, not all zero, with

Setting $p(x) = a_0 + a_1 x + \cdots + a_{n^2}x^{n^2}$, this is a nonzero polynomial (some $a_i \neq 0$) with $p(T) = 0$ and $\deg p \le n^2$. This proves Part 1.

Part 2 (Uniqueness). By Part 1, the set $I_T \setminus \{0\}$ of nonzero annihilating polynomials is nonempty. By the well-ordering principle applied to the (nonnegative integer) degrees of elements of $I_T \setminus \{0\}$, there is an element of minimal degree; dividing by its leading coefficient (nonzero, since $\mathbb{F}$ is a field) we obtain a monic polynomial $m$ of minimal degree in $I_T \setminus \{0\}$. This shows existence of a monic minimal-degree annihilator; call its degree $d$. Suppose $m$ and $m'$ are both monic of degree $d$ with $m(T) = m'(T) = 0$. Then $(m - m')(T) = 0$, and $\deg(m - m') < d$ because the degree-$d$ (leading) terms cancel (both are monic of the same degree $d$). If $m - m' \neq 0$, this contradicts minimality of $d$ among nonzero annihilators. Hence $m - m' = 0$, i.e., $m = m'$. This proves uniqueness; we call this polynomial $m_T$.

Part 3 (Divisibility). Let $g \in \mathbb{F}[x]$ with $g(T) = 0$. By the division algorithm in $\mathbb{F}[x]$ (valid since $\mathbb{F}$ is a field and $m_T$ has positive degree, as $m_T \neq 0$ has degree $d\geq 1$ — note $d \ge 1$ because the only degree-$0$ monic polynomial is the constant $1$, and $1(T) = I \neq 0$ since $V \neq \{0\}$... more precisely, if $V=\{0\}$ the whole theory is vacuous, so assume $\dim V \ge 1$, which we do), we may write

Applying the evaluation homomorphism at $T$ (which respects addition and multiplication of polynomials, since $T$ commutes with itself and scalars),

Since $g(T) = 0$ by hypothesis, we get $r(T) = 0$. If $r \neq 0$, then $r$ is a nonzero annihilator of $T$ with $\deg r < \deg m_T$, contradicting the minimality defining $m_T$. Hence $r = 0$, so $g(x) = q(x)\,m_T(x)$, i.e., $m_T \mid g$. Conversely, if $m_T \mid g$, write $g = q\, m_T$ for some $q \in \mathbb{F}[x]$; then $g(T) = q(T)\,m_T(T) = q(T)\cdot 0 = 0$. This proves the "if and only if." The claim that $m_T \mid \chi_T$ (the characteristic polynomial) will follow once we know $\chi_T(T) = 0$; that fact is the content of The Cayley-Hamilton Theorem, proved independently and used here only as a stated consequence, not as part of this proof. Granting Cayley–Hamilton, the divisibility $m_T \mid \chi_T$ is then an immediate application of the "only if" direction just proved, taking $g = \chi_T$.

Part 4 (Degree bound $\deg m_T \le n$). We give a direct proof not relying on Cayley–Hamilton, using triangularization. If $\mathbb{F}$ is algebraically closed (or more generally if $T$'s characteristic polynomial splits over $\mathbb{F}$), then by the theorem on Triangularization of Operators, there is a basis of $V$ in which $T$ is represented by an upper-triangular matrix $A$ with diagonal entries $\lambda_1, \dots, \lambda_n$ (the eigenvalues, with multiplicity). Consider $p(x) = (x - \lambda_1)(x-\lambda_2)\cdots(x - \lambda_n)$, a monic polynomial of degree $n$. We show $p(T) = 0$ directly. Let $v_1, \dots, v_n$ be the basis in which $T$ is upper triangular, so

for each $j$ (with the convention that the span is $\{0\}$ when $j=1$). We claim by induction on $j$ that

Base case $j=1$: $(T - \lambda_1 I) v_1 = 0$ by the triangularity relation above with $j=1$ (the span is trivial). Inductive step: suppose the claim holds for $j-1$, so $(T-\lambda_1I)\cdots(T-\lambda_{j-1}I)$ kills $v_1,\dots,v_{j-1}$. Since these operators commute (they are all polynomials in $T$), applying $(T-\lambda_1I)\cdots(T-\lambda_jI)$ to $v_k$ for $k < j$ still gives $0$ (the earlier factors already annihilate $v_k$, and applying the operator $(T - \lambda_j I)$ afterward to $0$ keeps it $0$). For $k = j$: write $(T-\lambda_j I)v_j = w \in \mathrm{span}(v_1,\dots,v_{j-1})$; then

by the inductive hypothesis applied to $w$, which is a linear combination of $v_1, \dots, v_{j-1}$, using linearity. This completes the induction. Taking $j = n$ shows $p(T) v_k = 0$ for every basis vector $v_k$, $k=1,\dots,n$, hence $p(T) = 0$ on all of $V$ by linearity. Thus $p$ is a monic annihilating polynomial of degree $n$, so by Part 3 (divisibility), $m_T \mid p$, giving $\deg m_T \le \deg p = n$. (If $\chi_T$ does not split over $\mathbb{F}$, one instead passes to a splitting field extension $\mathbb{K} \supseteq \mathbb{F}$; the matrix of $T$ triangularizes over $\mathbb{K}$, the same argument produces a degree-$n$ monic annihilator over $\mathbb{K}[x]$, and one checks the minimal polynomial computed over $\mathbb{F}$ agrees with that computed over $\mathbb{K}$ because the division algorithm argument in Part 3 works the same over either field and the annihilator ideal of a rational/integer matrix is generated by the same polynomial regardless of which field extension it is computed in — this is a standard field-extension invariance fact for annihilator ideals of matrices with entries in $\mathbb{F}$.) $\blacksquare$

($\Rightarrow$) Suppose $\lambda$ is an eigenvalue of $T$, with eigenvector $v \neq 0$, so $Tv = \lambda v$. Then for any polynomial $p$, an easy induction gives $T^k v = \lambda^k v$ for all $k \ge 0$ (base case $k=0$ trivial; inductive step $T^{k+1}v = T(T^k v) = T(\lambda^k v) = \lambda^k Tv = \lambda^{k+1}v$), and hence by linearity $p(T)v = p(\lambda) v$ for every polynomial $p$. Applying this to $p = m_T$:

Since $v \neq 0$, we conclude $m_T(\lambda) = 0$. ($\Leftarrow$) Suppose $m_T(\lambda) = 0$. By the factor theorem in $\mathbb{F}[x]$, write $m_T(x) = (x - \lambda) q(x)$ for some $q \in \mathbb{F}[x]$ with $\deg q = \deg m_T - 1$. Since $\deg q < \deg m_T$, minimality of $m_T$ forces $q(T) \neq 0$ (otherwise $q$ would be a smaller-degree nonzero annihilator, or $q=0$ which would force $m_T = 0$, impossible). Choose $w \in V$ with $q(T) w \neq 0$, and set $v = q(T) w \neq 0$. Then

using that $(T-\lambda I)$ and $q(T)$ commute and multiply to $m_T(T)$ as operators (polynomial evaluation is a ring homomorphism, so $(x-\lambda)q(x) = m_T(x)$ implies $(T - \lambda I) q(T) = m_T(T)$). So $Tv = \lambda v$ with $v \neq 0$, i.e., $\lambda$ is an eigenvalue. $\blacksquare$