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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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    $\begingroup$ By "diagonals", you mean the elements on the diagonal, right? Then this is always true, argue by induction, expanding the determinant by the first column. $\endgroup$ – астон вілла олоф мэллбэрг Feb 28 at 16:02
  • $\begingroup$ In infinite-order matrices the quantities may not exist. When they do, a straightforward application of the "permutation based" definition renders the two quantities equal. $\endgroup$ – Oscar Lanzi Feb 28 at 16:10
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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.

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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.

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  • $\begingroup$ Unless the matrices are of infinite order and the quantities do not exist. If they do exist, equality carries over to the infinite order case. $\endgroup$ – Oscar Lanzi Feb 28 at 16:14

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