# Thread: Linear independence and dependence

1. ## Linear independence and dependence

Hi,

I'm trying to understand linear independence from a graphical standpoint:

The reasoning behind why a is linear independent is because the vectors do not lie on the same plane when placed with their initial points at the origin - I struggling to see this. Can someone please explain?

b apparently is the opposite case (being linear dependent) but again, I fail to see the reasoning behind this.

- otownsend

2. ## Re: Linear independence and dependence

It's hard to tell just from a (rather small) picture but it looks to me like the three vectors in (b) all lie in a single plane.

3. ## Re: Linear independence and dependence

Is it reasonable for me to not see the observation you made? What specifically indicates in the picture that the three vectors all lie in a single plane? Do you mean that if you were to "take down" all the dimensions that the vectors would visually overlap?

4. ## Re: Linear independence and dependence

Originally Posted by otownsend
Is it reasonable for me to not see the observation you made? What specifically indicates in the picture that the three vectors all lie in a single plane? Do you mean that if you were to "take down" all the dimensions that the vectors would visually overlap?
Well of course it is reasonable! The diagrams are poor.
But if we imagine the $\nearrow \nwarrow$ shape can be slid (without any rotation) to just over the $\leftarrow$ shape then is appears that a triangle is formed. That determines a plane.

5. ## Re: Linear independence and dependence

Let's say the components of the vector are r,g,b.
For (a), you have vectors:
$\vec{v_1} = <-1,1,0>$
$\vec{v_2} = <1,1,0>$
$\vec{v_3} = <0,0,1>$
These are, indeed, linearly independent.

For (b), you have vectors:
$\vec{v_1} = <1,1,0>$
$\vec{v_2} = <1,0,1>$
$\vec{v_3} = <0,-1,1>$

$\vec{v_2} - \vec{v_1} = <1-1,0-1,1-0> = <0,-1,1> = \vec{v_3}$

They are not linearly independent.

6. ## Re: Linear independence and dependence

Originally Posted by SlipEternal
Let's say the components of the vector are r,g,b.
For (a), you have vectors:
$\vec{v_1} = <-1,1,0>$
$\vec{v_2} = <1,1,0>$
$\vec{v_3} = <0,0,1>$
These are, indeed, linearly independent.

For (b), you have vector
$\vec{v_1} = <1,1,0>$
$\vec{v_2} = <1,0,1>$
$\vec{v_3} = <0,-1,1>$

$\vec{v_2} - \vec{v_1} = <1-1,0-1,1-0> = <0,-1,1> = \vec{v_3}$

They are not linearly independent.
However,the question was about the given diagrams and not about particular vectors.

7. ## Re: Linear independence and dependence

Originally Posted by Plato
However,the question was about the given diagrams and not about particular vectors.
The question is about the particular vectors shown in the diagrams which can be given exact representations and shown to be linearly independent or linearly dependent.

8. ## Re: Linear independence and dependence

Originally Posted by SlipEternal
The question is about the particular vectors shown in the diagrams which can be given exact representations and shown to be linearly independent or linearly dependent.
You may be correct. But both HallsOfIvy & I read it otherwise.
Hi otownsend, what do you think?

9. ## Re: Linear independence and dependence

Well I can see if you collapsed the dimensions, that (b)'s vectors would overlap and therefore is linearly dependent. I guess that's a sufficient level of understand for this question