Product of Triangular Matrices

Theorem
Let $\mathbf A = \sqbrk a_n, \mathbf B = \sqbrk b_n$ be upper triangular matrices of order $n$.

Let $\mathbf C = \mathbf A \mathbf B$.

Then
 * $(1): \quad$ the diagonal elements of $\mathbf C$ are given by:
 * $\forall j \in \closedint 1 n: c_{j j} = a_{j j} b_{j j}$


 * That is, the diagonal elements of $\mathbf C$ are those of the factor matrices multiplied together.


 * $(2): \quad$ The matrix $\mathbf C$ is itself upper triangular.

The same applies if both $\mathbf A$ and $\mathbf B$ are lower triangular matrices.

Proof
From the definition of matrix product, we have:


 * $\ds \forall i, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} b_{k j}$

Now when $i = j$ (as on the main diagonal):
 * $\ds c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j}$

Now both $\mathbf A$ and $\mathbf B$ are upper triangular.

Thus:
 * if $k > j$, then $b_{k j} = 0$ and thus $a_{j k} b_{k j} = 0$
 * if $k < j$, then $a_{j k} = 0$ and thus $a_{j k} b_{k j} = 0$.

So $a_{j k} b_{k j} \ne 0$ only when $j = k$.

So:
 * $\ds c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j} = a_{j j} b_{j j}$

Now if $i > j$, it follows that either $a_{i k}$ or $b_{k j}$ is zero for all $k$, and thus $c_{i j} = 0$.

Thus $\mathbf C$ is upper triangular.

The same argument can be used for when $\mathbf A$ and $\mathbf B$ are both lower triangular matrices.