Product of Triangular Matrices

Theorem
Let $\mathbf A = \left[{a}\right]_n, \mathbf B = \left[{b}\right]_n$ be upper triangular matrices of order $n$.

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

Then:


 * The diagonal elements of $\mathbf C$ are given by: $\forall j \in \left[{1 .. n}\right]: c_{jj} = a_{jj} b_{jj}$.

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


 * 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:


 * $\displaystyle \forall i, j \in \left[{1 .. n}\right]: c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$

Now when $i=j$ (as on the diagonal), $\displaystyle c_{jj} = \sum_{k=1}^n a_{jk} b_{kj}$.

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

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

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

So $\displaystyle c_{jj} = \sum_{k=1}^n a_{jk} b_{kj} = a_{jj} b_{jj}$.

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

Thus $\mathbf C$ is upper triangular.

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