# Linear Independence of vectors question

1. Oct 9, 2012

### bossman007

1. The problem statement, all variables and given/known data

Suppose that A, B and C are not linearly independent. Then show how the $a_i$ can be computed, up to a common factor, from the scalar products of these vectors with each other

2. Relevant equations

$a_1$A + $a_2$B + $a_3$C = 0

$a_1$=$a_2$=$a_3$=0

Hint - Suppose that there are non-zero values of the $a_i$'s that satisfy
$a_1$A + $a_2$B + $a_3$C = 0. Then, taking the dot product of both sides of this equation with A will yield a set of equations that can be solved for the $a_i$'s

3. The attempt at a solution

$a_1$AA + $a_2$BA + $a_3$CA=0

no idea where to go from here, I took the dot product of both sides but confused from the wording of the question what my next step should be, or If I did my dot product right

2. Oct 9, 2012

### ehild

Do the same with B and C. The scalar product are numbers, so you have a system of equations for the three unknown parameters. As the vectors are not independent, one of the parameter can be chosen arbitrary. Solve the system of equations for the other two coefficients.

ehild

3. Oct 9, 2012

### bossman007

Thank you for your reply. you're saying arbitrarily choose a value? If so, I chose $a_1$=1

from that, my set of equations looks like, after moving the AA , AB and AC to the other side of the equation i get this:

$a_2$AB + $a_3$AC = -AA
$a_2$BB + $a_3$BC = -AB
$a_2$BC + $a_3$CC = -AC

**any vector combination above is a dot product, I just didnt know how to latex code it***

I went on to try substitution to solve for $a_2$, but the result was messy and didnt seem like I was on the right track. Am I on the right track?

4. Oct 10, 2012

### ehild

It will not be that messy. Multiply the first equation with BC, the second equation with AC. Subtract them. a3 cancels and you can isolate a2.

ehild