I am supposing that $A=(a_{ij})$ and $B=(b_{ij})$ are two $n\times n $ upper triangular square matrices. $\lambda \in \mathbb{R}$. So $a_{ij}=0$ whenever $i>j$.
I am trying to prove that these are also upper triangular, using definitions of matrix operations:
1) $AB$
2) $\lambda A$
3) $A+B$
For 1) I am doing something along the lines of:
$a_{i1}b_{1j}+\cdot\cdot\cdot+a_{in}b_{nj}$ (definition of matrix multiplication)
$i>j$ so I need to show that the expression evaluates to 0. But I am not sure how to approach it.
I am not sure on 2) or 3) either.
Best Answer
By definition,
$$AB=\left(\sum_{k=1}^na_{ik}b_{kj}\right)\;,\;\;\text{and for}\;\;i> j\;\;\;\sum_{k=1}^n a_{ik}b_{kj}=0\;,\;\;\text{since in any case}$$
$\;i>k\;\;or\;\;k>j\;$ , and thus the sum above is all zeros.
(2)-(3) are, imo, even easier:
$\;\lambda A=\left(\lambda a_{ij}\right)\;$, and clearly $\;\lambda a_{ij}=0\;$ for $\;i>j\;$ .
You try now $\;A+B=\left(a_{ij}+b_{ij}\right)\;$