Does adding linearly independent vectors retain linear independence?

Question

Suppose the vectors u, v, w are linearly independent and u'=u+v, v'=v+w and w'=u+w. I'd like to check if u', v', w' are also linearly independent.

I know they can be linearly independent, such as if u= (1,0,0), v=(0,1,0), w=(0,0,1). So, next I know that linear independence gives the smallest number of generators for a space, so u', v', w' could generate the space, however, I'm not sure how to prove this. How do I prove whether or not u', v', w' are linearly independent?

Any help to get me started would be appreciated.

Thank you.

Answer 1

Move $u, v, w$ to the standard vectors $e_1, e_2, e_3$ by an invertible linear transformation $A$, over the real numbers say. Then $e_1+e_2, e_2+e_3, e_1+e_3$ are independent and they are the image of $u+v,v+w,u+w$ by an invertible linear transformation. So the vectors $u',v',w'$ are also independent.

Answer 2

Hint/First step: suppose for some scalars $a,b,c$ we have

$$au' +bv' + cw' = 0$$

Given your expressions for $u',v',w'$ you can write the above as a combination of $u,v,w$ summing to zero. What does that tell you about the coefficients of that expression? and then what does that tell us about $a,b,c$

Answer 3

Though one can show your linear map is invertible by computing the determinant, there is another way, exploiting its circulant structure. Your linear map is $\, 1+S,\,$ where $\,S(u,v,w) = (v,w,u).\,$ Therefore $\,\color{#c00}{S^3 = 1}\,$ so $\,(1+S)(1-S+S^2) = 1+\color{#c00}{S^3} = 2,\,$ so $\,(1+S)^{-1} = \frac{1}2(1-S+S^2).$

Remark $\ $ This is a special case of pretty and practical circulant algebra.