Exercise 7, Section 5.3 of Hoffman’s Linear Algebra

Question

An $n\times n$ matrix $A$ is called triangular if $A_{ij} = 0$ whenever $i\gt j$ or if $A_{ij}=0$ whenever $i\lt j$. Prove that the determinant of a triangular matrix is the product $A_{11} A_{22}\dotsb A_{nn}$ of its diagonal entries.

My attempt: WLOG suppose $A$ is upper triangular matrix i.e. $A_{ij}=0$, if $i\gt j$. By theorem 2 section 5.3, $\text{det}(A)=\sum_{\sigma \in S_n}(\text{sgn}\sigma )A(1,\sigma(1))\dotsb A(n,\sigma(n))$. Then $(\text{sgn}\sigma )A(1,\sigma(1))\dotsb A(n,\sigma(n))\neq 0$, if $A(1,\sigma(1))\neq 0,…,A(n,\sigma(n))\neq 0$. Which implies $1\leq \sigma(1),…,n\leq \sigma(n)$. Since $n\leq \sigma(n)$, we have $n= \sigma(n)$. Since $\sigma$ is injective and $n-1\leq \sigma(n-1)$, we have $n-1= \sigma(n-1)$. Similarly $i=\sigma(i)$, $\forall i\in J_n$. So $\sigma=I$, identity map on $J_n$. Thus $\text{det}(A)$ $=\sum_{\sigma}(\text{sgn}\sigma )A(1,\sigma(1))\dotsb A(n,\sigma(n))$ $=(\text{sgn} I )A(1,1))\dotsb A(n,n)$ $= A(1,1))\dotsb A(n,n)$. Hence $\text{det}(A)= A(1,1))\dotsb A(n,n)$. Is my proof correct?

Potential approach: if $\sigma(i)\neq i$, for some $i\in J_n$, then show $A(k,\sigma(k))=0$, for some $k$. I can’t make any meaningful progress, because I think it’s really thought provoking.

Approach(3): Here is an alternative (but clever) solution, though it uses result from section 5.4. Which is, suppose we have an $n \times n$ matrix of the block form $\begin{bmatrix}A &B\\0&C\end{bmatrix}$, where $A\in M_{r\times r}(K)$, $C\in M_{s\times s}(K)$, $B\in M_{r\times s}(K)$ and $0\in M_{s\times r}(K)$, then $\text{det} \begin{bmatrix}A &B\\0&C\end{bmatrix}=\text{det}(A)\cdot \text{det}(C)$.

Answer 1

I can't follow your proof so I won't comment on whether or not it is correct, but I have never explicitly used this definition of the determinant to prove any of it's properties, and I always suggest you look for other ways first.

As to how I would solve the problem, I would just use induction on $n$. Indeed, let $U = \begin{bmatrix}a&x_{1\times(n-1)}\\0_{(n-1)\times1}&B_{(n-1)\times(n-1)}\end{bmatrix}$ be upper triangular, where $x$ is any vector, and $B$ is any upper triangular matrix. The determinant of $U$ is then $ det(U) = a*det(B)$, which you can see by expanding the determinant along the first column full of zeros (you can use the definition of the determinant here if you want to be fully transparent in your proof).