6

Consider the product $C=(C_{ij})$ of two upper triangular matrices $A=(a_{ij})$ and $B=(b_{ij})$, with $n=m$ (rows=columns). Deduce the expression of $C=(C_{ij})$.

I'm trying to do this proof but can't complete it.

Rodrigo de Azevedo
  • 23,223
Numbermind
  • 1,211
  • 1
    I think the definition says $C_{ij}=\sum_{k=1}^na_{ik}b_{kj}$. What kind of expression are you after? – awllower Jan 12 '17 at 17:55

2 Answers2

12

Definition. A $n \times n$ matrix $A$ is upper triangular if $a_{ij} = 0$ for $i >j$.

Now consider $A,B$ upper triangular and $C = AB$. Then consider $C_{ij}$ entry from $C$ with $i >j$, given by \begin{align} C_{ij} & = \sum_{k =1}^n a_{ik}b_{kj} \\ & = \sum_{k =1}^{i-1}a_{ik}b_{kj} + \sum_{k=i}^na_{ik}b_{kj} \\&= 0 + 0 = 0, \end{align} since \begin{align} k & \leq i-1 \implies a_{ik} =0 \\ k & \geq i \implies b_{kj} =0 \end{align} for this particular combination of $(i,j)$. We conclude that $C_{ij} = 0$ if $i >j$ hence $C$ is upper triangular.

We can easily generalize the statement we have proven to

Proposition. Let $U_1,\ldots,U_k$ be $n \times n$ upper triangular matrices. Then, $$V = \prod_{i=1}^k U_i$$ is an upper triangular matrix as well.

You can show it by means of prove by induction. Denote $P(k)$ the statement above. Then obviously $P(1)$ is true. Also $P(k) \implies P(k+1)$ (exercise!).

Gonçalo
  • 15,869
2

Let $A=[a_{ij}]$ and $B=[b_{ij}]$ be two $n \times n$ upper triangular matrices.

By definition

\begin{equation} a_{ij}=b_{ij}=0~~\text{if}~~~i>j. \end{equation}

Since $A^T, B^T$ are lower triangular matrices, and

\begin{equation} (AB)^T=B^TA^T \end{equation} is a lower trigular matrix (Ref. https://math.stackexchange.com/q/3881777), hence $AB$ is an upper triangular matrix. $\blacksquare$

mhmurad
  • 549