[proofplan]
The characteristic polynomial $p_A(\lambda) = \det(\lambda I - A)$ is a degree-$n$ polynomial with complex coefficients. By the Fundamental Theorem of Algebra, it has exactly $n$ roots in $\mathbb{C}$ counted with multiplicity, so it factors completely into linear factors over $\mathbb{C}$. Grouping repeated roots produces the stated factorisation.
[/proofplan]
[step:Establish that $p_A$ is a degree-$n$ polynomial over $\mathbb{C}$]
The characteristic polynomial of $A$ is defined by $p_A(\lambda) = \det(\lambda I - A)$. Expanding via the Leibniz formula,
\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 diagonal entry $(\lambda I - A)_{ii} = \lambda - a_{ii}$ is a degree-$1$ polynomial in $\lambda$, and each off-diagonal entry $(\lambda I - A)_{i,\sigma(i)} = -a_{i,\sigma(i)}$ is a constant. The identity permutation contributes $\prod_{i=1}^{n}(\lambda - a_{ii})$, which has degree $n$ with leading coefficient $1$. Every non-identity permutation $\sigma$ has at least two non-fixed points, so its product has degree at most $n - 2$. Therefore $p_A \in \mathbb{C}[\lambda]$ is a monic polynomial of degree $n$.
[/step]
[step:Apply the Fundamental Theorem of Algebra to factor $p_A$]
Since $p_A \in \mathbb{C}[\lambda]$ is a non-constant polynomial of degree $n \ge 1$, the [Fundamental Theorem of Algebra](/theorems/347) guarantees that $p_A$ has at least one root in $\mathbb{C}$. Applying the theorem inductively (factoring out a linear factor at each step), $p_A$ has exactly $n$ roots $\mu_1, \ldots, \mu_n \in \mathbb{C}$ counted with multiplicity, and since $p_A$ is monic,
\begin{align*}
p_A(\lambda) = (\lambda - \mu_1)(\lambda - \mu_2) \cdots (\lambda - \mu_n).
\end{align*}
[guided]
The [Fundamental Theorem of Algebra](/theorems/347) states that every non-constant polynomial in $\mathbb{C}[\lambda]$ has at least one root in $\mathbb{C}$. We apply this repeatedly: since $p_A$ has degree $n \ge 1$, it has a root $\mu_1 \in \mathbb{C}$. By the factor theorem, $(\lambda - \mu_1)$ divides $p_A(\lambda)$, giving $p_A(\lambda) = (\lambda - \mu_1) q_1(\lambda)$ where $q_1 \in \mathbb{C}[\lambda]$ has degree $n - 1$. If $n - 1 \ge 1$, apply the Fundamental Theorem of Algebra again to $q_1$ to find a root $\mu_2$. Continuing, after $n$ steps we obtain
\begin{align*}
p_A(\lambda) = (\lambda - \mu_1)(\lambda - \mu_2) \cdots (\lambda - \mu_n),
\end{align*}
since the leading coefficient of $p_A$ is $1$ (monic).
Why does this factorisation work over $\mathbb{C}$ but not necessarily over $\mathbb{R}$? Because the Fundamental Theorem of Algebra is specific to $\mathbb{C}$ — an irreducible polynomial over $\mathbb{R}$ can have degree $2$ (e.g., $\lambda^2 + 1$), but over $\mathbb{C}$ every irreducible polynomial has degree $1$.
[/guided]
[/step]
[step:Group repeated roots to obtain the stated factorisation]
Let $\lambda_1, \ldots, \lambda_k$ be the distinct values among $\mu_1, \ldots, \mu_n$, and for each $i = 1, \ldots, k$, define $m_a(\lambda_i) := |\{j : \mu_j = \lambda_i\}|$ to be the multiplicity of $\lambda_i$ as a root. Grouping the linear factors:
\begin{align*}
p_A(\lambda) = (\lambda - \lambda_1)^{m_a(\lambda_1)} \cdots (\lambda - \lambda_k)^{m_a(\lambda_k)}.
\end{align*}
Since every root $\mu_j$ belongs to exactly one group, the multiplicities sum to the total number of roots: $\sum_{i=1}^{k} m_a(\lambda_i) = n$.
[/step]
