# Basis for a subspace

1. May 24, 2014

### negation

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

Find the basis for the subspace 4x+y-3z

3. The attempt at a solution

I found that the basis is {[1;-4;0],[0;3;1]}. How do I know if it is linearly independent? I know that the mathematical definition of what LI is but how can it be applied to show in this case?

2. May 24, 2014

### HallsofIvy

Staff Emeritus
That is not a subspace- it is a linear expression! I suspect that you mean "Find a basis for the subspace of R3 of all (x, y, z) satisfying 4x+ y- 3z= 0".
(If you had written "Find a basis for the subspace 4x+ y- 3z= 0" I would have had no problem, interpreting it a short hand for the above. But the "= 0" is important. Also note "a basis" not "the basis". Any vector space or subspace has an infinite number of bases.)

The definition of "u, v are Linearly Independent" is that "if au+ bv= 0 then a= b= 0". Applied here, that would be a[1; -4; 0]+ b[0; 3; 1]= [a; -4a+ 3b; b]= [0; 0; 0] which tells you that a= 0; -4a+ 3b= 0; b= 0. The first and third equations tell you everything you need to know, don't they?

3. May 24, 2014

### negation

How do I know a[1;-4;0]+b[0;3;1] =[0;0;0]? Does it entails from the initial premise that
4x + y - 3z = 0?

secondly, we now know the basis B to be {[1;-4;0],[0;3;1]}

Suppose then [w]B = [-2;4], what then is [w]s?

This is a fairly simple question but the confusion stems from the fact that I am still fuzzy about how I should determine the direction of the change of coordinate of the basis B. From which coordinates to which does the basis b purports the movement to be?
If I know the basis B maps the coordinates from, for instance, a to b then I could set the equation up as [x]a = Bab[x]b

4. May 24, 2014

### LCKurtz

You are wanting to show those two vectors are linearly independent. What about Hall's explanation don't you understand?

5. May 24, 2014

### negation

Hi Kurtz,

That wasn't what I wanted. I read my notes and it says that the basis is linearly independent.
Where did the zero vector came from?

6. May 24, 2014

### LCKurtz

But you don't know it is a basis until you show the vectors are linearly independent. And that is what you asked, as quoted in red above. Halls explained how you show that, and his explanation, including the definition of linear independence, shows why you set the linear combination equal to zero.

7. May 24, 2014

### negation

I don't know what was I thinking but this part clicked. Thanks.

Second part, anyone?