Any vector masters?

[Priority]

Hey, im currently studying spesh for next year
And i seriously dont get the meaning of 'vector dependent'
Does it mean Vectors are parallel to each other?

[abeybaby]

could you give an example of what you mean? Do you mean linearly dependent/independent vectors?

[Priority]

Yeah, sorry im spesh noob:(

[keltingmeith]

Yeah, sorry im spesh noob:(

Hey man, don't stress it - we all gotta start somewhere!

Think of it like this - a set of vectors are linearly independent if you can't write one as a sum of other vectors. Let's start this a little simply:

The vectors i and j are independent, because there is no way of one to get the other. However, if I have the vectors a=i and b=2i, they are linearly independent, because b=2a.

But this might seem a bit trivial/obvious, so let's go up to three dimensions. Just like i and j are independent, i, j and k are also independent. Because no matter how many ways you try and express i and j together, you won't get k. Making a bit more sense? Let's take another step forward - say I have a=4i, b=2i-2j and c=i+j. Well, b+2c=2i-2j+2i+2j=4i=a, so a, b and c are linearly dependent because you can express one of the vectors in terms of the other 2.

So, let's jump to something that's NOT obvious at all:

a=2i-j
b=i+j
c=i-j

Are these three vectors linearly independent? Well, as we said above, a, b and c will be linearly independent if we CANNOT express one as the sum of two other vectors. In fact, if you think of linear dependency like this, it is actually easier to show if they're linearly dependent*, so let's do that: a=mb+nc, where m and n are some constants. So, expanding that out, we get:

2i-j=m(i+j)+n(i-j)=mi+mj+ni-nj

Now, here we can actually do something called "equate components", which is where we say that all the stuff in front of i is equal, and all the stuff in front of j is equal. This gives us:

2=m+n
-1=m-n

Solving these simultaneously, you should get:

2=n-1+n=2n-1 ===> n=3/2
m=2-n=1/2

So, this means that a=(1/2)b+(3/2)c
Since you can express one vector in terms of the other two, then all three vectors are linearly dependent of another.


*=note that there are multiple definitions of linear dependency, but they're all related in a way that they're basically the same test. However, you need to check for different things for different tests. For our test, things are dependent if only one solution exists. An equivalent test is to check la+mb+nc=0, and if the only solution is l=m=n=0, then the three vectors are independent. It doesn't matter which you use, just know which one you're using.

[Priority]

Omg thanks sir!,!,!,!
What an absolute legend:)
(So the way to know whether they r dependent/inde is to solve the two linear equations and sub in the the answers you got to the third equation)
Thanks again

[abeybaby]

Haha that's okay! Ok I'll try my best to explain (it's much easier in person with pretty diagrams). Also, apoplogies for the lengthy post.

Okay, lets define some things.

1. Unit vector
Any vector of magnitude one

2. vector i
A unit vector which points in the positive x direction

3. vector j
A unit vector which points in the positive y direction

4. vector k
A unit vector which points in the positive z direction
5. Linear combinations/span
The set of all possible vectors generated when a set of vectors are multiplied by real coefficients.

Okay, so let's think about some things. Suppose you're a particle living at the origin (0,0,0). let's say you’re sick of your boring 0 dimensional life and so you want to start to move.

consider you knew that the vector i exists, and you happen to like real numbers, so you let the number a be a real number. Let's see what all the possible places you could go would be.

you could go to (5,0,0), or (20,0,0) ot maybe (-7,0,0). You could go to any point whose coordinates are (a,0,0).

 But say you wanted to get out of this line, and go to other places. You simply can’t do it with only i, you need another vector. So then someone tells you about j, and no all of a sudden you can go to any place whose coordinates are (a, b, 0)! You can use some linear combination of i and j to get to (7,-3,0) or (6,4,0) or (-3,3,0) or ANYWHERE in the x-y plane.


So let’s step back and think for a second. You’re now very happy that you can live your particle existence in any place in the x-y plane, but now you’re greedy. You want more. You want to go EVEN FURTHER than your 2D existence. You weren’t happy with your 1D life, now you’re not happy with your 2D life, you want 3D. But what if no one ever told you about the vector k? How would you do it? Let’s think.

To get from 0D life as a point at (0,0,0), we took a vector, i, and took all the linear combinations of that vector, i.e., we took a*i and generated every point on a straight line. To get from 1D life to 2D life, we took all the possible linear combinations of 2 vector, i.e., a*i+b*j produced all of the 2D world. So if we want a 3D world, let’s take a 3rd vector and take all the linear combinations of those 3 vectors.
Let’s define our new vector to be something easy, let’s call it p and make it a nice simple 1i+1j. so let’s see where our new coordinate system takes us:

All the linear combinations of i, j, and p would be ai+bj+cp = (a+1)i+(b+1)j+0k. So we didn’t manage to produce a 3D world, the k component is still 0. Sad face. Diagrammatically: 



So p didn’t add any new information at all… but WHY! We followed all the steps! Pick a new vector ad take the linear combinations! The reason is that p can be expressed as a linear combination of i and j. That means that there is a way to write p as i*some scalar+j*some scalar, and so p is ALREADY INCLUDED in the information generated by the span of i and j. Therefore, we say that i, j and p are linearly dependent.

If we want to create a 3D world, we need to take the span of 3 linearly Independent vectors. Now the obvious choice is i, j, and k. But it doesn’t need to be that way. How about the three vectors i+j, i-j and i+j+k? What happens if we take the span of these three?



Now we can create 3D space! And if we took the span of the first 2, we’d create 2D space, because they are linearly independent. If we picked 3i as one vector and 10i for another, we would only manage 1D space since 3i and 10i are linearly dependent. So the dimension of the space created by the span depends on the number of linearly independent vectors, and that’s why we care whether or not a bunch of vectors are linearly dependent or not.

[wyzard]

Actually I've written some notes on this, check it out!