Upper triangular matrices product

Let $A \in \mathfrak{M}_{m \times n}(\mathbb{K})$ and $B \in \mathfrak{M}_{n \times p}(\mathbb{K})$ be two upper triangular matrices. We want to show that the product $AB$ is also an upper triangular matrix. So, let’s consider the $(i,j)$-th entry of the product $AB$:

$[AB]_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$

If we develop the expression above, we have:

$[AB]_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \ldots + a_{in}b_{nj}$

And for each $i > k$, we have $a_{ik} = 0$ because $A$ is upper triangular, and for each $j < k$, we have $b_{kj} = 0$ because $B$ is upper triangular. Therefore, each entry $[AB]_{ij}$ with $i > j$ is zero, which means that $AB$ is an upper triangular matrix.

The result for the product of lower triangular matrices is analogous, so we can conclude that the product of two upper (or lower) triangular matrices is an upper (or lower) triangular matrix.