Linear algebra proof

**ulysses123** · Mar 30th 2010, 03:53 AM

Let A be a m * p matrix whose columns all add to the same total s; and B be a p * n matrix whose columns all add to the same total t: Using summation notation prove that the n columns of AB each total st:

**hatsoff** · Mar 30th 2010, 04:21 AM

Originally Posted by ulysses123

Let A be a m * p matrix whose columns all add to the same total s; and B be a p * n matrix whose columns all add to the same total t: Using summation notation prove that the n columns of AB each total st:

By definition of matrix multiplication, the entries of $AB$ are given by $\sum_{i=1}^p a_{j,i} b_{i,k}$ .

We seek the sum of an arbitrary column of $AB$ , given by $\sum_{j=1}^m\left(\sum_{i=1}^p a_{j,i} b_{i,k}\right)$

$=\sum_{i=1}^p \left( b_{i,k} \sum_{j=1}^m a_{j,i}\right)$

$=\sum_{i=1}^p \left( b_{i,k} s\right)$

$=s\sum_{i=1}^p b_{i,k}$

$=st$ .

**ulysses123** · Mar 30th 2010, 03:24 PM

Thanks, what i was doing was expressing one column in Matrix A, by the sum of all columns divided by p, which was making it difficult to then simplify the expression for AB.

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