When is the determinant of a upper triangular not the product of its diagonals?
I'm trying to dive deeper into some topics for class, I was wondering if its possible to have an upper triangular matrix where its determinant is not equal to the product of its diagonals
Best Answer
Alternatively, we can also prove that for an upper triangular matrix $A$, we have $$ \det(A) = a_{11}a_{22}\dots a_{nn} $$ using just the definition of determinant.
Let $S_n$ be the set of all permutations of $\{1,2,\dots,n\}$. Recall that $$ \det(A) = \sum_{\sigma\in S_n} \text{sgn}(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}. $$ For any $\sigma\in S_n$ that is not the identity (i.e. $\sigma(i)\ne i$ for some $i\le n$), we let $j\in \{1,2,\dots,n\}$ be the largest number such that $\sigma(j)\ne j$. Since $j$ is largest, we must have $\sigma(j)< j$. This means that $a_{j\sigma(j)}=0$ since $A$ is upper triangular.
The above argument shows that $a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}=0$ for all other $\sigma\in S_n$ except the one that $\sigma(i)=i$ for all $i$. Since the sign of the identity permutation is $1$, this concludes what we want to prove.