Question

Linear Algebra: Show that the set is linearly independent. http://i46.tinypic.com/345enpi.jpg

Answer 1

If I write these as a system, is this a case where the constants must simply equal zero in order for the right hand side to equal the zero vector?

Answer 2

Help, I'm so lost

Answer 3

I'll give you a hint to get you started.

u1,u2, and u3 are linearly independent. By definition, this means:

c1*u1 + c2*u2 + c3*u3 = 0 implies that c1,c2, and c3 must all be 0.

Now, take u1+u2, u2+u3, and u3+u1 and write them in the same form. So,

k1*(u1+u2) + k2*(u2+u3) + k3*(u3+u1) = 0

The goal is to show k1, k2, and k3 are all 0. Then, you'd have linear independence. So, rearrange your terms until they match the form (something)*u1 + (something)*u2 + (something)*u3 = 0. Then, you know each of those something's are equal to 0. You'll have a system of equations, and solve.

Answer 4

OH!!! I had a problem like this involving polynomials and I had to factor out the variables......the coefficients (somethings) formed a system where each of the somethings had to equal zero in order to equal the zero vector on the RHS of each equation. I see it! Thank you!

Answer 5

You're welcome. If you want to double check your work, you're welcome to post your solution also. I don't mind double checking it for you.

Answer 6

ok, thanks

Answer 7

7. Suppose that the 3 vectors u₁, u₂, u₃ E Rⁿ are linearly independent. Show that the set {u₁ + u₂, u₂ + u₃, u₃ +u₁} is also linearly independent.

c₁(u₁+u₂) + c₂(u₂+u₃) + c₃(u₃+u₁) = 0
c₁u₁+ c₁u₂ + c₂u₂+ c₂u₃+ c₃u₃+ c₃u₁= 0
u₁(c₁+c₃) + u₂(c₁+c₂) + u₃(c₂+c₃) = 0
c₁+c₃=0
c₁+c₂=0
c₂+c₃=0
The system implies that c₁ = 0, c ₂ = 0, c ₃= 0. Therefore, the linear combination set {u₁ + u₂, u₂ + u₃, u₃ +u₁} is also linearly independent.

Answer 8

Yep. Good job.