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.

Start with the equation a(u+v)+b(v+w)+c(u+w)=0, and see if you can prove that a=b=c=0. — user84413, Jul 25 '14 at 22:41
Note that if the underlying field has characteristic $2$, we have linear dependence. — André Nicolas, Jul 25 '14 at 22:44

score 4 · Accepted Answer · answered Jul 25 '14 at 22:44

4

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.

answered Jul 25 '14 at 22:44

i. m. soloveichik

4,259

I'd like to verify your reply by restating it slightly differently. Essentially, if I show linear independence through creating a matrix of the initial vectors, I can simply do the vector addition above, which is nothing more than Gaussian operations, which means the vectors are still linearly independent. Does this make sense and is it correct? Thank you. – John Jul 25 '14 at 22:55
1

@Eric Yes, that's correct. – i. m. soloveichik Jul 25 '14 at 23:00
That's incredibly simple. Thank you for your time. – John Jul 25 '14 at 23:00

score 2 · Answer 2 · answered Jul 25 '14 at 22:43

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 $a,b,c$

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

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.

Does adding linearly independent vectors retain linear independence?

3 Answers3