Hi, i want to show that zi=cij*xj, where cij=aik*bkj, if yi=bij*xj and zi=aij*yj. (i,j,k are subscripts).

I have been struggling with this for a week. Any help is appreciated, thanks in advance.

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- Oct 9th 2013, 12:21 PMFilipVzLinear equations - Summation convention
Hi, i want to show that zi=cij*xj, where cij=aik*bkj, if yi=bij*xj and zi=aij*yj. (i,j,k are subscripts).

I have been struggling with this for a week. Any help is appreciated, thanks in advance. - Oct 9th 2013, 01:18 PMemakarovRe: Linear equations - Summation convention
In matrix form, you want to show that y = Bx and z = Ay implies that z = (AB)x. From the assumption, it follows that z = A(Bx), so what you need to show is associativity of matrix multiplication: A(Bx) = (AB)x.

Let A, B and C be matrices of compatible sizes. Then

$\displaystyle [AB]_{il}=\sum_k a_{ik}b_{kl}$ (1)

and

$\displaystyle [BC]_{kj}=\sum_l b_{kl}c_{lj}$ (2)

So we get

$\displaystyle \begin{align*}[(AB)C]_{ij} &= \sum_l\left(\sum_k a_{ik}b_{kl}\right)c_{lj} && \text{by (1)}\\ &=\sum_l\sum_k a_{ik}b_{kl}c_{lj}&& \text{by distributivity}\\ &=\sum_k\sum_l a_{ik}b_{kl}c_{lj}&& \text{by exchanging the order of summation, i.e., commutativity}\\ &=\sum_k a_{ik}\left(\sum_l b_{kl}c_{lj}\right)&& \text{by distributivity}\\ &=[A(BC)]_{ij}&& \text{by (2)} \end{align*}$ - Oct 9th 2013, 01:29 PMFilipVzRe: Linear equations - Summation convention
Hi, emakarov.

Thank you for the answer.

Is there any other way (more trivial) to prove that? - Oct 9th 2013, 01:51 PMemakarovRe: Linear equations - Summation convention
I am not sure. One can try writing the sums explicitly instead of using ∑. You can also try becoming more comfortable with ∑-notation and see what distributivity and exchanging limits means when written explicitly. Finally, you may search for explanations of the associativity of matrix multiplication; there must be good descriptions. (I am not sure if you realized that this problems boils down to associativity.)

Feel free to post here what you tried or found.