Question: I'm trying to practice proofs for my linear algebra final and I've been stuck on this one for some time. 

I have $AB = [A\mathbf{b\_1} \ A\mathbf{b\_2} \ \dots \ A\mathbf{b\_n}]$. I can show that $A\mathbf{b\_1}$ has the form $\left(
\begin{array}{c} a\_{11} b\_{11} \\
0 \\
\vdots \\
0
\end{array}
\right)$
and that $A\mathbf{b\_2}$ has the form $\left(
\begin{array}{c} a\_{11} b\_{12} + a\_{12}b\_{22} \\
a\_{22}b\_{22} \\
0 \\
\vdots \\
0
\end{array}
\right)$ but I'm not sure how to generalize this. Would proof by induction be appropriate?

Answer: 
Let $A=(a\_{ij})$ and $B=(b\_{ij})$. Since $B$ is upper triangular then

$$Be\_j=\sum\_{i=1}^{j} b\_{ij}e\_i$$
and since $A$ is also upper traingular then

$$AB e\_j=\sum\_{i=1}^jb\_{ij}Ae\_i=\sum\_{i=1}^jb\_{ij}\sum\_{k=1}^ia\_{ki}e\_k=\sum\_{k=1}^j\sum\_{i=k}^jb\_{ij}a\_{ki}e\_k=\sum\_{k=1}^jc\_{kj}e\_k$$
which means that $AB$ is upper triangular.
Denote the $ij$-th ($i$-th row, $j$-th column) element of $A$ by $A\_{ij}$, we have that $A\_{ij}=0$ if $i>j$. The same holds for $B$ with the notation $B\_{ij}$

Now if $i>j$:

$$(AB)\_{ij}=\sum\_{k=1}^{n}A\_{ik}B\_{kj}=\sum\_{k=i}^{n}A\_{ik}B\_{kj}$$

But if $k≥i>j$ then $k>j$ and hence $B\_{kj}=0$. This gives $(AB)\_{ij}=0$ for $i>j$, i.e. $AB$ is upper triangular.