Prove Linear Independence of Vector Sums

[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?

 

[Office_Shredder]

Linearly independent from what?

 

[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.

 

[Office_Shredder]

You can't. Try to find the linear dependency

 

[canephalanx]

I see. How can I prove otherwise?

 

[Office_Shredder]

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

 

[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?

 

[Mark44]

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.

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

 

[HallsofIvy]

$$\vec{v_{1}}, \vec{v_{1}}, \mbox{ and }\vec{v_1}+\vec{v_2}$$ 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?

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?