# Linear Independence

1. Sep 16, 2010

### canephalanx

How can I prove given an arbitrary set of vectors v1 and v2, given they are linearly independent, that their sum (v1 + v2) is also linearly independent?

2. Sep 16, 2010

### Office_Shredder

Staff Emeritus
Linearly independent from what?

3. Sep 16, 2010

### canephalanx

let me rephrase that, how can i show that v1,v2, and v1+v2 is linearly independent given v1 and v2 is linearly independent. I seemed to have left out a key statement.

4. Sep 16, 2010

### Office_Shredder

Staff Emeritus
You can't. Try to find the linear dependency

5. Sep 16, 2010

### canephalanx

I see. How can I prove otherwise?

6. Sep 17, 2010

### Office_Shredder

Staff Emeritus
Well, how do you prove that a set of vectors is linearly dependent?

7. Sep 17, 2010

### JThompson

$$\vec{v_{1}}, \vec{v_{1}}, \mbox{ and }\vec{v_1}+\vec{v_2}$$ are linearly independent if the only solution to

$$a\vec{v_{1}}+b\vec{v_{2}}+c(\vec{v_{1}}+\vec{v_{2}})=0$$

is $$a=0, b=0, c=0$$.

Is this the case, or can you find other values that satisfy this equation?

8. Sep 17, 2010

### Staff: Mentor

The set {v1, v1, v1 + v2 } is always linearly dependent, since the third one listed is a linear combination of the first two.

9. Sep 17, 2010

### HallsofIvy

Well, that reduces to (a+ c)v1+ (b+ c)v1= 0. Since v1 and v2 are independent, you must have a+ c= 0 and b+ c= 0. Obviously a= b= c is one solution to that but those are only two equations in three unknows. We can typically solve two equations in two unknowns. Okay, solve for a and b, say, treating c as a number. Then let c be whatever you want.

Solving for a and b "in terms of c" gives a= -c and b= -c. Take c to be anything you like and find a and b. If you happen to select c= 0, then, of course, you get a= b= c= 0. But what if you select c= 1?