# Thread: need help with proof

1. ## need help with proof

A= [aij] n x n lower triangular matrix
B= [bij] n x n lower triangular matrix
C= [cij] = AB

Proof cij= 0 for i < j or lower diagonal

cij= [ ai1b1j + ai2b2j + ...+ai(j-1)b(j-1)j ] + [ aijbjj +....+ainbnj ]

I am a little confuse of what they are trying to show within the 2 brackets.

2. Originally Posted by justin016
A= [aij] n x n lower triangular matrix
B= [bij] n x n lower triangular matrix
C= [cij] = AB

Proof cij= 0 for i < j or lower diagonal

cij= [ ai1b1j + ai2b2j + ...+ai(j-1)b(j-1)j ] + [ aijbjj +....+ainbnj ]

I am a little confuse of what they are trying to show within the 2 brackets.
Let $A = (a_{ij})$ be a lower triangular $n\times n$ matrix, this means $a_{ij} = 0$ for $1\leq i.
Let $B = (b_{ij})$ be a lower triangular $n\times n$ matrix, this means $b_{ij} = 0$ for $1\leq i < j \leq n$.
Define, $C = AB = (c_{ij})$, then $c_{ij} = \Sigma_k a_{ik}b_{kj}$.
Say that $i then $a_{ik}b_{kj}=0$ for all $k=1,2,...,n$ so $c_{ij}=0$*.

*)If $k then $b_{kj}=0$ otherwise $i and then $a_{ik}=0$.

btw, is this thought process correct:
If we multiply a matrix A by a matrix B, assuse dimensions are so that
multiplication is defined, then AB=C, and the columns of C
will be a linear combination of the columns of A, using the the elements in
column j, j=1,....,n as weights ?

4. thanx hacker for the reply

if k is the column of a, how can k be greater than itself? in the statement

i < j < k?

5. Originally Posted by justin016
I consider two cases: (i) $k or (ii) $j\leq k$.
In case (i) since $k it means $b_{kj} = 0$.
In case (ii) since $j\leq k$ and $i it implies $i and so $a_{ik}=0$.
Therefore, $a_{ik}b_{kj}=0$ for $k=1,2,...,n$ thus $c_{ij}=\Sigma_k a_{ik}b_{kj} = 0$ for $i.