# Statements Let $A\in \mathbb{C}^{n\times n}$. The [[Characteristic polynomial of linear operator|characteristic polynomial]] $c_{A}(\lambda)=\det(A-\lambda I_{n})$ satisfies $c_{A}(A)=0_{n}$. We can replace $\mathbb{C}$ with any algebraically closed field. # Proofs ###### General proof Recall that $c_{A}(x)$ is the determinant of $B:= A-xI_{n}$. The characteristic property of the [[Adjugate of square matrix|adjugate matrix]] yields $\text{adj}(A-xI_{n}) (A-xI_{n})= c_{A}(x) I_{n} \tag{1}.$ Let $P(x)=\text{adj}(A-xI_{n})$ and $Q(x)=A-xI_{n}$ which are polynomials in indeterminate $x$ with matrix coefficients, i.e. $P, Q\in \text{Mat}_{n \times n}(\mathbb{C})[x]$. Then the LHS of $(1)$ equals $(P\cdot Q)(x)$ where $P\cdot Q$ denotes the Cauchy product of the polynomials. In general, it may not be the case that the matrix product $P(M)Q(M)$ is the same as $(P\cdot Q)(M)$. However, this holds if $M$ commutes with coefficients of $Q$: distributivity yields $P(M)\,Q(M) \;=\; \sum_{j,k} P_j\,M^j \,Q_k\,M^k = \sum_{j,k} P_j\,Q_k\,M^{j+k} \;=\; (P\cdot Q)(M).$In our case, the coefficients of $Q$, which are $A$ and $-I_{n}$, commute with $A$. Thus setting $M=A$ and $c_{A}(A) = 0_{n}$ yields $c_{A}(A)=0_{n}.$ ###### Diagonalisable case We will show that for any $p\in \mathbb{C}[x]$, all eigenvalues of $p(A)$ have the form $p(\lambda)$ where $\lambda$ are eigenvalues of $A$. Let $\mu$ be an eigenvalue of $p(A)$. The [[Fundamental Theorem of Algebra|fundamental theorem algebra]] allows us to factorise $p(x)-\mu$ as follows $p(x) - \mu = \gamma \prod_{i=1}^{s} (x- \delta_{j})$where $\gamma,\delta_{j}\in \mathbb{C}$. Since $\det(p(A)-\mu I_{n})=0$, the multiplicativity of the determinant yields $\det(A-\delta_{j}I_{n})=0$ for some $1\leq j\leq s$. Moreover, evaluation at $x=\delta_{j}$ yields $p(\delta_{j})=\mu$ . This shows that all eigenvalues of $p(A)$ have the form $p(\lambda)$ where $\lambda$ are eigenvalues of $A$. Now suppose $A$ has distinct eigenvalues $\lambda_{1},\dots,\lambda_{n}$. Then $A=P^{-1}DP$ where $D=\text{diag}(\lambda_{1},\lambda_{2},\dots,\lambda_{n})$ . Furthermore, $A^{i} = P^{-1}DP$ yields $c_{A}(A)=P^{-1}c_{A}(D)P$ but $c_{A}(D)=0_{n,n}$ since $c_{A}(\lambda_{i})=0$ for $i=1,2,\dots,n$ ($\lambda_{i}$ are eigenvalues of $A$). This shows that $c_{A}(A)=0$. # Applications ###### Calculating the minimal polynomial of matrix over algebraically closed field The Cayley–Hamilton theorem tells us that the characteristic polynomial $c_A(\lambda)$ annihilates the matrix $A$, so the [[Minimal polynomial of square matrix|minimal polynomial]] $\mu_A(\lambda)$ must divide $c_A(\lambda).$ Meanwhile, the fact that [[Eigenvalues are roots of minimal polynomial|every root of the characteristic polynomial in the ground field (eigenvalues) is a root of the minimal polynomial]] implies that for each eigenvalue $\lambda_i$, the factor $(\lambda - \lambda_i)$ must appear in $\mu_A(\lambda)$ with **at least** exponent $1$. Since $\mu_A$ divides $c_A$, and $c_A$ contains each eigenvalue $\lambda_i$ with some multiplicity $m_i$, the exponent of $(\lambda - \lambda_i)$ in $\mu_A(\lambda)$ must be between $1$ and $m_i$. That is: $\mu_A(\lambda) = \prod_{i} (\lambda - \lambda_i)^{r_i}, \quad \text{with } 1 \le r_i \le m_i.$