How to find the non-trivial linear combination of "A,B,C" that adds up to 0 after finding out the vectors are Linearly dependent.

Question

A = [24] B = [-4] C = [-4]

[-18] [5] [2]

[-4], [0] [1]

Answer 1

Hey Reece,

Lots of big important concepts here.

Let's start with linear dependence, just in case you're not strong there. Linear dependence means that there exist scalars (let's call them a, b, and c, but some people will use k's or lambdas) such that:

aA + bB +cC = 0 where a, b, and c are not all 0.

So how do we find the a, b, and c scalars that make this a true statement? Let's start by writing our vectors as the columns of a matrix (M)

M =[24 -4 -4]

[-18 5 2]

[-4 0 1]

(Hopefully that's all lined up for you.)

The next big idea that's helpful here is that row operations don't change the DEPENDENCY RELATIONSHIP of the column space. What does that mean? Well let's get the matrix into reduced row echelon form first. I'll assume you know how to do that, but we should get:

[1 0 -0.25]

[0 1 -0.5 ]

[0 0 0 ]

In this form, it's clear to see that our third column is simply -0.25 times our first column plus -0.5 times our second column. The beauty here is that this same dependency relationship still applies to our original columns!

So we have that our vector C = -0.25A -0.5B. You can actually do the math there to confirm this is the case (actually try it). Well now we can use some basic algebra. Let's get everything over to the left side and we'll have:

0.25A + 0.5B + C = 0. So we've found non zero scalars (namely 0.25, 0.5, and 1) that create a nontrivial linear combination of A, B, and C that results in 0.

Hope that helps! Feel free to reach out with any questions.