[proofplan] We compute the determinant of $tI_n-A$ directly from the Leibniz formula. The matrix $tI_n-A$ has the same triangular shape as $A$, and its diagonal entries are $t-a_{ii}$. In the Leibniz expansion of a triangular matrix, every non-identity permutation contributes zero, so the determinant is exactly the product of the diagonal entries. [/proofplan]

[step:Pass from $A$ to the triangular matrix $tI_n-A$] Let $R:=k[t]$, and regard $tI_n-A$ as a matrix in $M_n(R)$. Define $B=(b_{ij})\in M_n(R)$ by \begin{align*} B:=tI_n-A. \end{align*} Thus, for each $i,j\in\{1,\dots,n\}$, \begin{align*} b_{ij}=t\delta_{ij}-a_{ij}, \end{align*} where $\delta_{ij}\in k$ denotes the Kronecker delta. If $A$ is upper triangular, then $a_{ij}=0$ whenever $i>j$. For such $i,j$, also $\delta_{ij}=0$, so $b_{ij}=0$. Hence $B$ is upper triangular. If $A$ is lower triangular, then $a_{ij}=0$ whenever $i<j$, and the same computation gives $b_{ij}=0$ whenever $i<j$. Hence $B$ is lower triangular. In both cases, the diagonal entries of $B$ are \begin{align*} b_{ii}=t-a_{ii} \end{align*} for $i\in\{1,\dots,n\}$. [/step]

[step:Show that only the identity permutation contributes to the determinant]Let $S_n$ denote the symmetric group of all permutations of $\{1,\dots,n\}$. For $\sigma\in S_n$, let $\operatorname{sgn}(\sigma)\in R$ denote the image in $R=k[t]$ of the integer sign of the permutation $\sigma$. By the Leibniz formula, which defines the determinant over the commutative ring $R$, \begin{align*} \det B=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n b_{i,\sigma(i)}. \end{align*} Assume first that $B$ is upper triangular. If a term indexed by $\sigma\in S_n$ is nonzero, then $b_{i,\sigma(i)}\ne 0$ for every $i\in\{1,\dots,n\}$, so upper triangularity forces $\sigma(i)\ge i$ for every $i$. Summing these inequalities gives \begin{align*} \sum_{i=1}^n \sigma(i)\ge \sum_{i=1}^n i. \end{align*} Since $\sigma$ is a permutation of $\{1,\dots,n\}$, the two sums are equal. Therefore every inequality $\sigma(i)\ge i$ is an equality, so $\sigma(i)=i$ for every $i$. Thus $\sigma=\operatorname{id}$. Assume now that $B$ is lower triangular. If a term indexed by $\sigma\in S_n$ is nonzero, then lower triangularity forces $\sigma(i)\le i$ for every $i$. Summing gives \begin{align*} \sum_{i=1}^n \sigma(i)\le \sum_{i=1}^n i. \end{align*} Again the two sums are equal because $\sigma$ is a permutation, so $\sigma(i)=i$ for every $i$. Thus $\sigma=\operatorname{id}$. Consequently, in either triangular case, the only nonzero term in the Leibniz formula is the term indexed by the identity permutation.[/step]

[guided]Recall that $R=k[t]$ and that $B=(b_{ij})\in M_n(R)$ is the matrix $B=tI_n-A$. The determinant is a sum over all ways of choosing one entry from each row and one entry from each column. The triangular shape makes almost all of those choices impossible: if a chosen entry falls on the zero side of the triangle, the whole product becomes zero. Let $S_n$ be the group of permutations of $\{1,\dots,n\}$. For $\sigma\in S_n$, let $\operatorname{sgn}(\sigma)\in R$ denote the image in $R$ of the integer sign of $\sigma$. The Leibniz formula, used as the determinant definition in the [polynomial ring](/page/Polynomial%20Ring) $R=k[t]$, says \begin{align*} \det B=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n b_{i,\sigma(i)}. \end{align*} Here a permutation $\sigma$ chooses the entry $b_{i,\sigma(i)}$ from row $i$, and the product uses exactly one entry from each column because $\sigma$ is bijective. Suppose first that $B$ is upper triangular. Then $b_{ij}=0$ whenever $i>j$. Therefore, if \begin{align*} \prod_{i=1}^n b_{i,\sigma(i)} \end{align*} is nonzero, none of its factors may lie below the diagonal. For every $i\in\{1,\dots,n\}$, this forces $\sigma(i)\ge i$. Summing over $i$ gives \begin{align*} \sum_{i=1}^n \sigma(i)\ge \sum_{i=1}^n i. \end{align*} But $\sigma$ is a permutation, so the list $\sigma(1),\dots,\sigma(n)$ is just a reordering of $1,\dots,n$. Hence \begin{align*} \sum_{i=1}^n \sigma(i)=\sum_{i=1}^n i. \end{align*} The only way a finite collection of inequalities $\sigma(i)\ge i$ can have equality after summing is that each individual inequality is equality. Thus $\sigma(i)=i$ for every $i$, so $\sigma=\operatorname{id}$. We now repeat the argument for the lower triangular case with the inequality reversed. If $B$ is lower triangular, then $b_{ij}=0$ whenever $i<j$. A nonzero determinant term must therefore satisfy $\sigma(i)\le i$ for every $i$. Summing gives \begin{align*} \sum_{i=1}^n \sigma(i)\le \sum_{i=1}^n i. \end{align*} Since $\sigma$ is still a permutation, the two sums are equal, and therefore every inequality $\sigma(i)\le i$ is equality. Hence $\sigma=\operatorname{id}$. Thus, whether $B$ is upper triangular or lower triangular, every non-identity permutation contributes a zero product to the determinant expansion.[/guided]

[step:Evaluate the surviving determinant term] Since only the identity permutation contributes, and $\operatorname{sgn}(\operatorname{id})=1$, the determinant formula reduces to \begin{align*} \det B=\prod_{i=1}^n b_{ii}. \end{align*} Using $b_{ii}=t-a_{ii}$ for each $i\in\{1,\dots,n\}$, we obtain \begin{align*} \det(tI_n-A)=\prod_{i=1}^n(t-a_{ii}). \end{align*} By the defining convention $\chi_A(t)=\det(tI_n-A)$, this gives \begin{align*} \chi_A(t)=\prod_{i=1}^n(t-a_{ii}). \end{align*} This is the desired formula. [/step]