Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an old exam question I remember how to do, but not why it makes sense. Take a matrix $A$:

$A=\left(\begin{array}{rrrr} 1 & 6 & 2 & -4 \\ -3 & 2 & -2 & -8 \\ 4 & -1 & 3 & 9 \end{array}\right) $

Then row reduce it:

$\text{rref}(A)=\left(\begin{array}{rrrr} 1 & 0 & \frac{4}{5} & 2 \\ 0 & 1 & \frac{1}{5} & -1 \\ 0 & 0 & 0 & 0 \end{array}\right) $

But then if you want the rightmost (in this case) linearly dependent vector from $A$ expressed as linear combination of column vectors from $A$, you can use the rightmost column of rref$(A)$ as weights;

$\left(\begin{array}{rrr} 1 & 6 & 2 \\ -3 & 2 & -2 \\ 4 & -1 & 3 \end{array}\right)$$\left(\begin{array}{r} 2 \\ -1 \\ 0 \end{array}\right)= \left(\begin{array}{r} -4 \\ -8 \\ 9 \end{array}\right)$

And you can use the second rightmost column of rref$(A)$ as weights if you want to express the second dependent column vector as a linear combination of column vectors from $A$.

Edit: With 'weights' I mean viewing the matrix multiplication above as a linear combination of the column vectors, where the vector you multiply with contains the weights, a la $2\bf{v_1}-1\bf{v_2}+0\bf{v_3}$, where $\bf{v_1,v_2,v_3}$ are the first three column vectors of $A$.

I've been trying to remember why this is obvious (is it obvious?) but I'm not getting anywhere. Thanks for any replies.

share|improve this question
add comment

1 Answer 1

up vote 0 down vote accepted

You can view

$A=\left(\begin{array}{rrrr} 1 & 6 & 2 & -4 \\ -3 & 2 & -2 & -8 \\ 4 & -1 & 3 & 9 \end{array}\right) $

As an augmented matrix $A\vec{x}=\vec{y}$, with $\vec{y}={(-4,-8,9)}^T$. The the row reduction then solves for exactly the weights needed to create $\vec{y}$ out of $\bf{v_1,v_2,v_3}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.