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.
Multiplication of two upper triangular matrices
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linear-algebra
matrices
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1I think the definition says $C_{ij}=\sum_{k=1}^na_{ik}b_{kj}$. What kind of expression are you after? – 2017-01-12
1 Answers
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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_{ki}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 wel.
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!).