Linear Algebra - independent vectors spanning R^3

  • #1

Homework Statement


Suppose that S={u,v,w} is a basis of R3.
a) Is {u-v,u+v} linearly independent? Why or why not?
b) Does {u+v,u-v} span R3? Why or why not?


Homework Equations


na


The Attempt at a Solution


a) Because we know that u,v,w are linearly independent and span R3, c1u+c2v+c3w = 0.
To test u+v, u-v, consider a(u+v)+b(u-v)=0. So, au+av+bu-bv=0, so u(a+b)+v(a-b)=0. Since u and v are linearly independent, a+b=0 and a-b=0. So 2b=0, so b=0, so a=0. So u+v, u-v are linearly independent.
b) Because there are only 2 vectors, u and v, the rank can only be 2, so it cannot span R3.

I'm pretty sure I got a) correct, but I'm not sure about b). Thanks!
 

Answers and Replies

  • #2
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Homework Statement


Suppose that S={u,v,w} is a basis of R3.
a) Is {u-v,u+v} linearly independent? Why or why not?
b) Does {u+v,u-v} span R3? Why or why not?


Homework Equations


na


The Attempt at a Solution


a) Because we know that u,v,w are linearly independent and span R3, c1u+c2v+c3w = 0.
To test u+v, u-v, consider a(u+v)+b(u-v)=0. So, au+av+bu-bv=0, so u(a+b)+v(a-b)=0. Since u and v are linearly independent, a+b=0 and a-b=0. So 2b=0, so b=0, so a=0. So u+v, u-v are linearly independent.
b) Because there are only 2 vectors, u and v, the rank can only be 2, so it cannot span R3.

I'm pretty sure I got a) correct, but I'm not sure about b). Thanks!
Both a and b are right.
 
  • #3
vela
Staff Emeritus
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You got the first one right. The second one is essentially correct, but your terminology is off. When you say "rank", you're referring to a matrix or a linear transformation; here you have just two vectors. The span of {u+v, u-v} is a subspace of dimension 2.
 

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