# When is the determinant of a upper triangular not the product of its diagonals?

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

• 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. – астон вілла олоф мэллбэрг Feb 28 at 16:02
• 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. – Oscar Lanzi Feb 28 at 16:10

## 2 Answers

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.

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.

• 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. – Oscar Lanzi Feb 28 at 16:14