Key idea: The adjugate matrix $\operatorname{adj}(\lambda I - A)$ is a matrix whose entries are polynomials in $\lambda$, and the identity $(\lambda I - A) \cdot \operatorname{adj}(\lambda I - A) = \det(\lambda I - A) \cdot I$ will give us the result when we "substitute" $\lambda = A$.

Step 1: The adjugate identity. For any matrix $M$, we have $M \cdot \operatorname{adj}(M) = \det(M) \cdot I$. Applying this to $M = \lambda I - A$:

$(\lambda I - A) \cdot \operatorname{adj}(\lambda I - A) = p_A(\lambda) \cdot I.$

Step 2: Expand the adjugate as a polynomial in $\lambda$. Each entry of $\operatorname{adj}(\lambda I - A)$ is an $(n-1) \times (n-1)$ minor of $\lambda I - A$, hence a polynomial in $\lambda$ of degree at most $n - 1$. So we can write:

$\operatorname{adj}(\lambda I - A) = B_{n-1}\lambda^{n-1} + B_{n-2}\lambda^{n-2} + \cdots + B_1 \lambda + B_0,$

where $B_0, B_1, \ldots, B_{n-1}$ are $n \times n$ matrices with entries in $F$ (independent of $\lambda$).

Step 3: Expand both sides. Write $p_A(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0$. The left side is:

$(\lambda I - A)(B_{n-1}\lambda^{n-1} + \cdots + B_0) = B_{n-1}\lambda^n + (B_{n-2} - AB_{n-1})\lambda^{n-1} + \cdots + (B_0 - AB_1)\lambda + (-AB_0).$

The right side is:

$p_A(\lambda) \cdot I = I\lambda^n + c_{n-1}I\lambda^{n-1} + \cdots + c_1 I \lambda + c_0 I.$

Step 4: Match coefficients. Comparing coefficients of each power of $\lambda$:

$B_{n-1} = I, \quad B_{n-2} - AB_{n-1} = c_{n-1}I, \quad \ldots, \quad B_0 - AB_1 = c_1 I, \quad -AB_0 = c_0 I.$

Step 5: Multiply and sum. Multiply the $k$-th equation (coefficient of $\lambda^k$) by $A^k$ on the left:

• $\lambda^n$: $A^n \cdot B_{n-1} = A^n \cdot I = A^n$.
• $\lambda^{n-1}$: $A^{n-1}(B_{n-2} - AB_{n-1}) = c_{n-1} A^{n-1}$.
• $\vdots$
• $\lambda^1$: $A(B_0 - AB_1) = c_1 A$.
• $\lambda^0$: $-AB_0 = c_0 I$.

Wait, we need to be more careful. Multiply the equation for $\lambda^k$ by $A^k$:

• From $\lambda^n$: $B_{n-1} = I$ $\implies$ multiply by $A^n$: $A^n B_{n-1} = A^n$.
• From $\lambda^{n-1}$: $B_{n-2} - AB_{n-1} = c_{n-1}I$ $\implies$ multiply by $A^{n-1}$: $A^{n-1}B_{n-2} - A^n B_{n-1} = c_{n-1}A^{n-1}$.
• From $\lambda^{n-2}$: $B_{n-3} - AB_{n-2} = c_{n-2}I$ $\implies$ multiply by $A^{n-2}$: $A^{n-2}B_{n-3} - A^{n-1}B_{n-2} = c_{n-2}A^{n-2}$.
• $\vdots$
• From $\lambda^0$: $-AB_0 = c_0 I$ $\implies$ multiply by $I$: $-AB_0 = c_0 I$.

Step 6: Telescope. Adding all these equations, the left side telescopes:

$A^n B_{n-1} + (A^{n-1}B_{n-2} - A^n B_{n-1}) + \cdots + (-AB_0) = -AB_0 + \text{(everything cancels)}.$

More precisely, the sum is $-AB_0$ from the last equation, but the telescoping gives:

$\text{LHS} = -AB_0 + AB_0 - A^2 B_1 + A^2 B_1 - \cdots = 0.$

Wait, let me re-sum. The total of the left sides is:

$A^n - A^n + A^{n-1}B_{n-2} - A^{n-1}B_{n-2} + \cdots = 0.$

Actually, the positive term $A^k B_{k-1}$ from the $\lambda^k$ equation cancels with the negative term $-A^k B_{k-1}$ from the $\lambda^{k-1}$ equation. So the total is $0$.

The total of the right sides is $A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I = p_A(A)$.

Therefore $p_A(A) = 0$. $\blacksquare$

Special case: Assume $A$ is diagonalizable, i.e., $A = PDP^{-1}$ where $D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$.

Then $p_A(\lambda) = \prod_{i=1}^n (\lambda - \lambda_i)$, and:

$p_A(A) = \prod_{i=1}^n (A - \lambda_i I) = P \prod_{i=1}^n (D - \lambda_i I) P^{-1}.$

Now $D - \lambda_i I = \operatorname{diag}(\lambda_1 - \lambda_i, \ldots, \lambda_n - \lambda_i)$, which has a zero in the $i$-th diagonal entry. The product $\prod_i (D - \lambda_i I)$ has every diagonal entry equal to zero (the $j$-th entry is $\prod_i (\lambda_j - \lambda_i)$, which contains the factor $\lambda_j - \lambda_j = 0$). So $\prod_i (D - \lambda_i I) = 0$, hence $p_A(A) = 0$.

Over an algebraically closed field $F$, let $\lambda$ be an eigenvalue of $A$ with eigenvector $v$. Then:

$p_A(A)v = p_A(\lambda) v = 0 \cdot v = 0,$

since $A^k v = \lambda^k v$ for all $k$, so $p(A)v = p(\lambda)v$ for any polynomial $p$.

If $A$ has $n$ linearly independent eigenvectors $v_1, \ldots, v_n$ (i.e., $A$ is diagonalizable), then $p_A(A)v_i = 0$ for all $i$ implies $p_A(A) = 0$.

For non-diagonalizable $A$: use generalized eigenvectors. If $v$ is a generalized eigenvector with $(A - \lambda I)^m v = 0$, and $p_A(\lambda) = (\lambda - \lambda_0)^{m_0} \cdots$, then a similar but more involved argument shows $p_A(A)v = 0$. Since the generalized eigenvectors span $F^n$ (over an algebraically closed field), $p_A(A) = 0$.