Product of Triangular Matrices

Theorem
Let $$\mathbf{A} = \left[{a}\right]_{n}, \mathbf{B} = \left[{b}\right]_{n}$$ be a 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:

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

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

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

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.