Product matrix as a Linear Combination

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SUMMARY

The discussion centers on expressing the product of a row vector y and a matrix A as a linear combination of the row matrices of A. The participant successfully derives that yA can be represented as y1[a11, a12, ..., a1n] + y2[a21, a22, ..., a2n] + ... + ym[am1, am2, ..., amn]. This formulation highlights the contribution of each element of y as a scalar coefficient for the respective rows of A. The solution emphasizes the importance of recognizing the structure of matrix multiplication in linear algebra.

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Saladsamurai
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Problem Statement

Let

[tex]\mathbf{y} = [y_1\, y_2\, ...\, y_m][/tex]

And

[tex]A =<br /> \left[\begin{array} {cccc}<br /> a_{11}&a_{12}&...&a_{1n}\\<br /> a_{21}&a_{22}&...&a_{2n}\\<br /> a_{m1}&a_{m2}&...&a_{mn}<br /> \end{array}\right] [/tex]

Show that the product yA can be expressed as a linear combination of the row matrices of A
with the scalar coefficients coming from y



Attempt at Solution

I thought that I would write out the actual product, which is a row vector. I thought that
something might jump out at me from here:

yA = [(y1a11 + y2a21 + ... + ymam1) (y1a12 + y2a22 + ... + ymam2) (y1a1n + y2a2n + ... + ymamn)]

I am not sure where to go from here. I know that it is going to be a summation of the rows of A ... but what I have now is just written column-wise... and it is not a summation.

A hint maybe?
 
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I think that you have a row vector. You can treat it just like a column vector and break it down into a sum. For example (a + b + c, d + b, b + c) = (a, d, b) + (b, b, c) + (c, 0, 0).

Pull out terms that have the same factor from y.
 
I think that I got it!

I just wrote the summation of the rows of A :

[a11 a12 ... a1n] + [a21 a22 ... a2n] + ... + [am1 am2 ... amn]

and then noted that each term needs to multiplied by each of the elements of y :

y1[a11 a12 ... a1n] + y2[a21 a22 ... a2n] + ... + ym[am1 am2 ... amn]

I guess I just thought the solution would have been a little more 'graceful' as opposed to 'guess and check.'

thanks!
 

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