How to prove using Leibniz formula that the determinant of a upper triangular matrix is equal to product of its diagonals?

Question

Let F be a field, and n a positive integer. Using the Leibniz formula for the determinant, prove that the determinant of an upper triangular matrix $U ∈ F^{n×n}$ or a lower triangular matrix $L ∈ F^{n×n}$ is equal to the product of its diagonal entries.

So I'm sort of confused as to what to do? I was going to create an arbitrary matrix and then show by cofactor expansion what the determinant was equal to and then start again with an arbitrary matrix and then create a triangle in bottom left or upper right of the matrix. Then take a product of the diagonal but I'm unsure how to create this triangle? Any thoughts as to how to prove this?

Answer 1

Essentially the idea is to show that only the identity permutation can produce a nonzero product.

To see this, let $\sigma$ be a nonidentity permutation. Then there must be some $k$ with $\sigma(k)<k$. But then the product $\prod_{i=1}^na_{i\sigma(i)}$ contains $a_{k\sigma(k)}$ which lies below the diagonal and hence the product is zero.

Then the determinant is necessarily the product of the diagonal entries (corresponding to the identity permutation) as you desire.

Edit: this is for upper triangular matrices. For lower triangular matrix either use the transpose or adjust the argument accordingly.

Answer 2

Notice that if $n = 1$, then the determinant of an upper triangular matrix is indeed the product of its (only) diagonal entry.

Now let $U = (U_{ij}) \in F^{n\times n}$ be an upper triangular matrix, and suppose the result is true for an upper triangular matrix of dimension $n-1$. Using the cofactor expansion of the determinant with the first column, $$ \det{U} = U_{11} M_{11} + \sum_{i=2}^n U_{i1} M_{i1}, $$ where $M_{i1}$ is the $(i,1)$ minor of $U$. But since $U$ is upper triangular, $U_{i1} = 0$ for every $i > 1$, so $\det{U} = U_{11} M_{11}$. Now $M_{11}$ is the determinant of the matrix formed by removing the first row and the first column of $U$, so it is an upper triangular matrix of dimension $n-1$. By the induction hypothesis, its determinant is the product of its diagonal entries, that is $M_{11} = \prod_{i=2}^n U_{ii}$. So $$ \det{U} = U_{11} \prod_{i=2}^n U_{ii} = \prod_{i=1}^n U_{ii}. $$ Since the result if true for $n=1$, then it is true for any dimension $n$.

The same goes for a lower triangular matrix, but it can also be seen from the upper triangular case since the determinant of matrix is equal to the determinant of its transposed.