[proofplan] We define $p_A(\lambda) = \det(\lambda I - A)$ and expand using the Leibniz formula over $S_n$. The identity permutation contributes a degree-$n$ product $\prod_{i}(\lambda - A_{ii})$ whose leading terms give the $\lambda^n$ and $-(\operatorname{tr} A)\lambda^{n-1}$ coefficients; every non-identity permutation contributes degree at most $n - 2$, so it cannot affect these two leading coefficients. The constant term is read off by evaluating at $\lambda = 0$. When the polynomial splits over an algebraically closed field, comparing coefficients of the factored form $\prod_i (\lambda - \lambda_i)$ with the Leibniz expansion yields the trace and determinant identities. [/proofplan] [step:Expand $p_A(\lambda) = \det(\lambda I - A)$ via the Leibniz formula] Define $p_A(\lambda) := \det(\lambda I - A)$. The Leibniz formula for the determinant gives \begin{align*} p_A(\lambda) &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^{n} (\lambda I - A)_{i,\sigma(i)}. \end{align*} Each entry of $\lambda I - A$ is \begin{align*} (\lambda I - A)_{i,j} &= \begin{cases} \lambda - A_{ii} & \text{if } j = i, \\ -A_{ij} & \text{if } j \neq i. \end{cases} \end{align*} For a permutation $\sigma \in S_n$, let $\operatorname{Fix}(\sigma) := \{i \in \{1, \ldots, n\} : \sigma(i) = i\}$ denote its set of fixed points. Each fixed point $i \in \operatorname{Fix}(\sigma)$ contributes a factor $\lambda - A_{ii}$ (degree 1 in $\lambda$), while each non-fixed point $i \notin \operatorname{Fix}(\sigma)$ contributes $-A_{i,\sigma(i)}$ (degree 0). Therefore the product $\prod_{i=1}^{n} (\lambda I - A)_{i,\sigma(i)}$ is a polynomial in $\lambda$ of degree $|\operatorname{Fix}(\sigma)|$, and $p_A(\lambda)$ is a polynomial of degree at most $n$. [guided] We begin from the definition $p_A(\lambda) := \det(\lambda I - A)$. To extract the coefficients we need a concrete expansion of this determinant. The Leibniz formula expresses any $n \times n$ determinant as a sum over all permutations $\sigma \in S_n$: \begin{align*} p_A(\lambda) &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^{n} (\lambda I - A)_{i,\sigma(i)}. \end{align*} The matrix $\lambda I - A$ has entries \begin{align*} (\lambda I - A)_{i,j} &= \begin{cases} \lambda - A_{ii} & \text{if } j = i, \\ -A_{ij} & \text{if } j \neq i. \end{cases} \end{align*} This means that in the product $\prod_{i=1}^{n} (\lambda I - A)_{i,\sigma(i)}$, each factor is either $\lambda - A_{ii}$ (when $\sigma$ fixes $i$) or $-A_{i,\sigma(i)}$ (a constant in $\lambda$, when $\sigma$ moves $i$). Let $\operatorname{Fix}(\sigma) := \{i : \sigma(i) = i\}$. Then the product has exactly $|\operatorname{Fix}(\sigma)|$ factors that are degree 1 in $\lambda$, and $n - |\operatorname{Fix}(\sigma)|$ factors that are degree 0. So the product is a polynomial of degree $|\operatorname{Fix}(\sigma)|$. Since every permutation contributes a polynomial in $\lambda$, $p_A(\lambda)$ is itself a polynomial. What is its degree? The identity permutation $\sigma = \operatorname{id}$ fixes all $n$ points, contributing a degree-$n$ term. Every other permutation fixes at most $n - 2$ points (a non-identity permutation moves at least 2 elements), so it contributes degree at most $n - 2$. Therefore $p_A$ has degree exactly $n$. [/guided] [/step] [step:Extract the leading coefficient and the $\lambda^{n-1}$ coefficient from the identity permutation] The identity permutation $\sigma = \operatorname{id}$ contributes \begin{align*} \prod_{i=1}^{n} (\lambda - A_{ii}) &= \lambda^n - \Bigl(\sum_{i=1}^{n} A_{ii}\Bigr) \lambda^{n-1} + \text{(terms of degree } \leq n-2\text{)}. \end{align*} The coefficient of $\lambda^n$ is $1$, and the coefficient of $\lambda^{n-1}$ is $-\sum_{i=1}^{n} A_{ii} = -\operatorname{tr} A$. Every non-identity permutation $\sigma \neq \operatorname{id}$ has at least two non-fixed points, so $|\operatorname{Fix}(\sigma)| \leq n - 2$, and the corresponding product has degree at most $n - 2$. Therefore no non-identity permutation can contribute to the $\lambda^n$ or $\lambda^{n-1}$ terms. It follows that \begin{align*} p_A(\lambda) &= \lambda^n - (\operatorname{tr} A)\, \lambda^{n-1} + \text{(terms of degree } \leq n-2\text{)}. \end{align*} [guided] We now isolate the two leading coefficients. The identity permutation $\sigma = \operatorname{id}$ has $\operatorname{Fix}(\operatorname{id}) = \{1, \ldots, n\}$ and $\operatorname{sgn}(\operatorname{id}) = +1$, so it contributes \begin{align*} \prod_{i=1}^{n} (\lambda - A_{ii}). \end{align*} Expanding this product: each factor is $(\lambda - A_{ii})$, and the leading term (choosing $\lambda$ from every factor) is $\lambda^n$. To get $\lambda^{n-1}$, we choose $-A_{ii}$ from exactly one factor and $\lambda$ from the remaining $n - 1$ factors. Summing over the choice of which factor contributes $-A_{ii}$: \begin{align*} \text{coefficient of } \lambda^{n-1} &= \sum_{i=1}^{n} (-A_{ii}) = -\sum_{i=1}^{n} A_{ii} = -\operatorname{tr} A. \end{align*} Can any other permutation affect these coefficients? If $\sigma \neq \operatorname{id}$, then $\sigma$ moves at least one element $i$, so $\sigma(i) \neq i$. But permutations decompose into disjoint cycles, and a single non-fixed point must belong to a cycle of length $\geq 2$, so $\sigma$ moves at least 2 elements. This means $|\operatorname{Fix}(\sigma)| \leq n - 2$, and the product $\prod_i (\lambda I - A)_{i,\sigma(i)}$ has degree at most $n - 2$ in $\lambda$. Therefore non-identity permutations contribute nothing to the coefficients of $\lambda^n$ or $\lambda^{n-1}$. Combining: \begin{align*} p_A(\lambda) &= \lambda^n - (\operatorname{tr} A)\, \lambda^{n-1} + \text{(lower-order terms)}. \end{align*} [/guided] [/step] [step:Evaluate $p_A(0) = (-1)^n \det A$ to identify the constant term] Setting $\lambda = 0$ in the definition $p_A(\lambda) = \det(\lambda I - A)$: \begin{align*} p_A(0) &= \det(0 \cdot I - A) = \det(-A) = (-1)^n \det A, \end{align*} where the last equality uses the multilinearity of the determinant: scaling all $n$ rows by $-1$ multiplies the determinant by $(-1)^n$, as given by the [Properties of the Determinant](/theorems/917). [/step] [step:Combine to write $p_A(\lambda) = \lambda^n - (\operatorname{tr} A)\,\lambda^{n-1} + \cdots + (-1)^n \det A$] From the previous steps, $p_A$ is a monic polynomial of degree $n$ with: - Leading coefficient $1$ (coefficient of $\lambda^n$), - Coefficient of $\lambda^{n-1}$ equal to $-\operatorname{tr} A$, - Constant term $p_A(0) = (-1)^n \det A$. Therefore \begin{align*} p_A(\lambda) &= \lambda^n - (\operatorname{tr} A)\,\lambda^{n-1} + \cdots + (-1)^n \det A, \end{align*} where "$\cdots$" stands for the intermediate coefficients of degrees $n - 2$ down to $1$. [/step] [step:Factor over an algebraically closed field and compare coefficients to obtain the eigenvalue identities] Now assume $F$ is algebraically closed (e.g., $F = \mathbb{C}$). By the [Characteristic Polynomial Splits over $\mathbb{C}$](/theorems/3278), $p_A$ factors completely as \begin{align*} p_A(\lambda) &= (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n), \end{align*} where $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$ counted with algebraic multiplicity (these are the roots of $p_A$ by the [Eigenvalues as Roots of the Characteristic Polynomial](/theorems/918)). **Constant term:** Evaluating at $\lambda = 0$: \begin{align*} (-1)^n \det A = p_A(0) = (-\lambda_1)(-\lambda_2) \cdots (-\lambda_n) = (-1)^n \prod_{i=1}^{n} \lambda_i. \end{align*} Dividing both sides by $(-1)^n$: \begin{align*} \det A &= \prod_{i=1}^{n} \lambda_i. \end{align*} **Coefficient of $\lambda^{n-1}$:** Expanding the product $\prod_{i=1}^{n} (\lambda - \lambda_i)$, the coefficient of $\lambda^{n-1}$ is $-\sum_{i=1}^{n} \lambda_i$ (choosing $-\lambda_i$ from exactly one factor and $\lambda$ from the rest). Equating with the coefficient found above: \begin{align*} -\operatorname{tr} A = -\sum_{i=1}^{n} \lambda_i, \qquad \text{hence} \qquad \operatorname{tr} A = \sum_{i=1}^{n} \lambda_i. \end{align*} [guided] When $F$ is algebraically closed, every polynomial of degree $n \geq 1$ in $F[\lambda]$ splits into linear factors. By the [Characteristic Polynomial Splits over $\mathbb{C}$](/theorems/3278), the characteristic polynomial factors as \begin{align*} p_A(\lambda) &= (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n), \end{align*} where $\lambda_1, \ldots, \lambda_n \in F$ are the eigenvalues of $A$ (repeated according to algebraic multiplicity). These are eigenvalues because they are exactly the roots of $p_A$, and by the [Eigenvalues as Roots of the Characteristic Polynomial](/theorems/918), $\lambda$ is an eigenvalue of $A$ if and only if $p_A(\lambda) = 0$. We now compare coefficients between the Leibniz expansion $p_A(\lambda) = \lambda^n - (\operatorname{tr} A)\lambda^{n-1} + \cdots + (-1)^n \det A$ and the factored form. **Determinant as product of eigenvalues.** Set $\lambda = 0$ in the factored form: \begin{align*} p_A(0) &= (0 - \lambda_1)(0 - \lambda_2) \cdots (0 - \lambda_n) = (-1)^n \lambda_1 \lambda_2 \cdots \lambda_n. \end{align*} We already computed $p_A(0) = (-1)^n \det A$. Equating and dividing by $(-1)^n$: \begin{align*} \det A &= \lambda_1 \lambda_2 \cdots \lambda_n = \prod_{i=1}^{n} \lambda_i. \end{align*} **Trace as sum of eigenvalues.** Expand the factored form: \begin{align*} \prod_{i=1}^{n} (\lambda - \lambda_i) &= \lambda^n - \Bigl(\sum_{i=1}^{n} \lambda_i\Bigr) \lambda^{n-1} + \text{(lower-order terms)}. \end{align*} The coefficient of $\lambda^{n-1}$ in $\prod_i (\lambda - \lambda_i)$ is $-\sum_i \lambda_i$, by the same expansion argument used for $\prod_i (\lambda - A_{ii})$ earlier (choose $-\lambda_i$ from one factor, $\lambda$ from all others). From the Leibniz expansion, the coefficient of $\lambda^{n-1}$ in $p_A(\lambda)$ is $-\operatorname{tr} A$. Equating: \begin{align*} -\operatorname{tr} A &= -\sum_{i=1}^{n} \lambda_i, \qquad \text{so} \qquad \operatorname{tr} A = \sum_{i=1}^{n} \lambda_i. \end{align*} These two identities — determinant as product, trace as sum — are the Vieta relations for the characteristic polynomial. They hold over any algebraically closed field, and in particular over $\mathbb{C}$. [/guided] [/step]