[proofplan] We work with the adjugate of $tI - A$ in the polynomial matrix ring $\mathrm{Mat}_n(\mathbb{F}[t])$. The [Adjugate Identity](/theorems/397) gives $(tI - A) \cdot \mathrm{adj}(tI - A) = \chi_A(t) \cdot I$. Expanding both sides in powers of $t$ and equating coefficients yields $n + 1$ matrix equations. Multiplying the $t^k$ equation by $A^k$ on the left and summing produces a telescoping cancellation on the left side, leaving $\chi_A(A) = 0$ on the right. [/proofplan] [step:Write the adjugate as a polynomial in $t$ with matrix coefficients] Let $A \in \mathrm{Mat}_n(\mathbb{F})$ and consider $tI - A \in \mathrm{Mat}_n(\mathbb{F}[t])$. The adjugate $\mathrm{adj}(tI - A)$ has entries that are cofactors of $tI - A$, hence polynomials in $t$ of degree at most $n - 1$. Write \begin{align*} \mathrm{adj}(tI - A) = B_{n-1} t^{n-1} + B_{n-2} t^{n-2} + \cdots + B_1 t + B_0, \end{align*} where $B_0, \dots, B_{n-1} \in \mathrm{Mat}_n(\mathbb{F})$ are constant matrices. [/step] [step:Apply the adjugate identity and equate coefficients] By the [Adjugate Identity](/theorems/397) in $\mathrm{Mat}_n(\mathbb{F}[t])$: \begin{align*} (tI - A) \cdot \mathrm{adj}(tI - A) = \chi_A(t) \cdot I. \end{align*} Write $\chi_A(t) = t^n + c_{n-1}t^{n-1} + \cdots + c_1 t + c_0$. Expanding the left side and equating coefficients of $t^k$ on both sides: \begin{align*} t^n &: \quad B_{n-1} = I, \\ t^{n-1} &: \quad B_{n-2} - AB_{n-1} = c_{n-1} I, \\ t^{n-2} &: \quad B_{n-3} - AB_{n-2} = c_{n-2} I, \\ &\;\;\vdots \\ t^1 &: \quad B_0 - AB_1 = c_1 I, \\ t^0 &: \quad -AB_0 = c_0 I. \end{align*} [guided] The expansion of the left side uses the distributive law. The term $(tI - A)(B_{n-1}t^{n-1} + \cdots + B_0)$ produces $B_{n-1}t^n$ from the $tI$ factor hitting $B_{n-1}t^{n-1}$, and $-AB_0$ from the $-A$ factor hitting $B_0$. In general, the coefficient of $t^k$ for $1 \leq k \leq n - 1$ is $B_{k-1} - AB_k$, coming from $tI \cdot B_{k-1}t^{k-1}$ and $(-A) \cdot B_k t^k$. Matching these with the right side $\chi_A(t) \cdot I$ gives the system of $n + 1$ matrix equations listed above. [/guided] [/step] [step:Multiply each equation by $A^k$ and sum to obtain $\chi_A(A) = 0$] Multiply the equation for $t^k$ on the left by $A^k$: \begin{align*} A^n &: \quad A^n B_{n-1} = A^n, \\ A^{n-1} &: \quad A^{n-1}B_{n-2} - A^n B_{n-1} = c_{n-1} A^{n-1}, \\ A^{n-2} &: \quad A^{n-2}B_{n-3} - A^{n-1}B_{n-2} = c_{n-2} A^{n-2}, \\ &\;\;\vdots \\ A^1 &: \quad A B_0 - A^2 B_1 = c_1 A, \\ A^0 &: \quad -AB_0 = c_0 I. \end{align*} Summing all equations, the left side telescopes: each term $A^k B_{k-1}$ appears with $+$ from the $(k)$th equation and $-A^k B_{k-1}$ from the $(k+1)$th equation, so all intermediate terms cancel, leaving $0$. The right side sums to \begin{align*} A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I = \chi_A(A). \end{align*} Therefore $\chi_A(A) = 0$. [guided] The telescoping is the heart of the argument. Write out the left sides in order from top to bottom: \begin{align*} &A^n B_{n-1} \\ + \; &A^{n-1}B_{n-2} - A^n B_{n-1} \\ + \; &A^{n-2}B_{n-3} - A^{n-1}B_{n-2} \\ &\;\;\vdots \\ + \; &A B_0 - A^2 B_1 \\ + \; &(-AB_0). \end{align*} Reading column by column: $A^n B_{n-1}$ appears in row 1 with $+$ and in row 2 with $-$, cancelling. Similarly $A^{n-1}B_{n-2}$ cancels between rows 2 and 3, and so on down to $AB_0$ cancelling between the last two rows, so the total is $0$. Meanwhile the right sides sum to $A^n + c_{n-1}A^{n-1} + \cdots + c_0 I = \chi_A(A)$, giving $0 = \chi_A(A)$. Note that one cannot "substitute $t = A$" directly into $(tI - A)\,\mathrm{adj}(tI - A) = \chi_A(t)\,I$ because $A$ is a matrix, not a scalar, and the expressions are polynomials in $t$ with matrix coefficients. The coefficient-matching-and-summing procedure avoids this issue. [/guided] [/step] [step:Extend to endomorphisms and deduce $M_\alpha \mid \chi_\alpha$] Since the [characteristic polynomial is a similarity invariant](/theorems/402), $\chi_A$ depends only on the endomorphism $\alpha$ represented by $A$. The identity $\chi_A(A) = 0$ holds for any matrix representing $\alpha$, so $\chi_\alpha(\alpha) = 0$. By [Existence and Uniqueness of Minimal Polynomial](/theorems/405), $M_\alpha(t) \mid \chi_\alpha(t)$. Since $\deg \chi_\alpha = \dim V$, this gives $\deg M_\alpha \leq \dim V$. [/step]