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
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No, it is not possible.
Assuming that $A$ is an $n\times n$ upper (or lower) triangular matrix whose diagonal entries are $d_1,\dots, d_n$.
We can perform a series of row operations on $A$ which consists entirely of adding a multiple of one row to another row. This operation does not change the determinant of $A$.
If we have $d_i=0$ for some $i\le n$, then the row operations would give us a row which consists entirely of zeroes, hence $\det(A) = 0 = d_1 d_2\dots d_n$.
If we $d_i\ne 0$ for all $i\le n$, then the row operations would give us a diagonal matrix whose elements on the diagonal are still $d_1,\dots,d_n$. Obviously, the determinant of the diagonal matrix is $d_1 d_2\dots d_n$, thus we have $$ \det(A) = d_1 d_2\dots d_n $$ in either cases.
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.