[proofplan] ($\Rightarrow$) If $\alpha$ is upper triangular, $\chi_\alpha(t) = \prod(t - \lambda_i)$ where the $\lambda_i$ are diagonal entries. ($\Leftarrow$) If $\chi_\alpha$ splits, induction on $\dim V$: extract an eigenvalue from the splitting, find an eigenvector, pass to the quotient $V/\langle v_1 \rangle$ where the induced endomorphism still has a splitting characteristic polynomial, and lift the resulting triangular form back to $V$. [/proofplan] [step:Show that an upper triangular matrix has a splitting characteristic polynomial] Suppose $\alpha$ is represented by an upper triangular matrix $T$ with diagonal entries $\lambda_1, \dots, \lambda_n$. Then $tI - T$ is also upper triangular with diagonal entries $t - \lambda_1, \dots, t - \lambda_n$. The determinant of an upper triangular matrix is the product of its diagonal entries: \begin{align*} \chi_\alpha(t) = \det(tI - T) = \prod_{i=1}^n (t - \lambda_i), \end{align*} which splits into linear factors over $\mathbb{F}$. [/step] [step:Extract an eigenvalue and eigenvector when $\chi_\alpha$ splits] Suppose $\chi_\alpha(t) = \prod_{i=1}^n (t - \lambda_i)$ splits over $\mathbb{F}$. In particular, $\lambda_1$ is a root of $\chi_\alpha$, so $\det(\lambda_1 I - A) = 0$. By the [Determinant Invertibility Criterion](/theorems/396), $\alpha - \lambda_1\,\mathrm{id}$ is not invertible, hence $\ker(\alpha - \lambda_1\,\mathrm{id}) \neq \{0\}$. Choose an eigenvector $v_1 \neq \mathbf{0}$ with $\alpha(v_1) = \lambda_1 v_1$. [/step] [step:Pass to the quotient and verify the characteristic polynomial still splits] Let $W = \langle v_1 \rangle$. Since $\alpha(v_1) = \lambda_1 v_1 \in W$, the subspace $W$ is $\alpha$-invariant. The quotient endomorphism $\bar{\alpha}: V/W \to V/W$ is well-defined by $\bar{\alpha}(v + W) = \alpha(v) + W$. Extend $v_1$ to a basis $(v_1, v_2, \dots, v_n)$ of $V$. In this basis, the matrix of $\alpha$ has the block form $\begin{pmatrix} \lambda_1 & * \\ \mathbf{0} & C \end{pmatrix}$ where $C$ is the matrix of $\bar{\alpha}$ with respect to $(v_2 + W, \dots, v_n + W)$. By the [Block Triangular Determinant](/theorems/399): \begin{align*} \chi_\alpha(t) = (t - \lambda_1)\,\det(tI_{n-1} - C) = (t - \lambda_1)\,\chi_{\bar{\alpha}}(t). \end{align*} Since $\chi_\alpha(t) = \prod_{i=1}^n (t - \lambda_i)$, dividing gives $\chi_{\bar{\alpha}}(t) = \prod_{i=2}^n (t - \lambda_i)$, which splits. [/step] [step:Complete the induction to construct an upper triangular basis] **Base case $n = 1$:** every $1 \times 1$ matrix is upper triangular. **Inductive step:** We have $\dim(V/W) = n - 1$ and $\chi_{\bar{\alpha}}$ splits. By the inductive hypothesis, there exists a basis of $V/W$ in which $\bar{\alpha}$ is upper triangular. Lift this basis to vectors $v_2, \dots, v_n \in V$ and prepend $v_1$. In the basis $(v_1, v_2, \dots, v_n)$, the matrix of $\alpha$ is upper triangular: the first column has $\lambda_1$ on the diagonal and zeros below (since $\alpha(v_1) = \lambda_1v_1$), and the lower-right $(n-1) \times (n-1)$ block is upper triangular by the inductive hypothesis. [/step]