# $(v_1,v_2\ldots,v_n)$ is linearly independent. Can $(v_1+u_1,v_2+u_2,\ldots,v_n+u_n)$ remain linearly independent?

Given are $n$-dimensional vector space $\langle R^n;+\mid R\rangle$ and $n$ vectors $(v_1,v_2,\ldots,v_n)$ which are linearly independent, $u_i\in R^n, i=\overline{1,n}.$

Say, $x$ = $(v_1,v_2,\ldots,v_n)$ and $y$ = $(u_1,u_2,\ldots,u_n)$. Then $x+y\in R^n$. Vectors are both $n$-dimensional.

Q: So, by that way, are the $n$ vectors in $x+y$ linearly independent? Is there some theorem which relates to this?

• sorry, linearly independent from what? Or do you mean a basis of $n$ vectors $v_1,...,v_n$? – AnalysisStudent0414 May 7 '14 at 17:10
• I guess it follows. – MoDzeus May 7 '14 at 17:11
• @Modestas_S are you assuming that the $u_i$ are also linearly independent? – essay May 7 '14 at 17:15

The answer is no, even with the extra hypothesis $u_i$ linearly independent
Take for example $u_i = -v_i$. They both are sets of independent vectors, but $x+y=(0,\dots,0)$
Consider, in the real vector space $\mathbb{R}^2$, $v_1 = (1,0)$ and $v_2 = (0,1)$, which are linearly independent. Consider aldo $u_1 = (0,1)$ and $u_2 = (1,0)$. Clearly the vectors $v_1 + u_1$ and $v_2 + u_2$ are not linearly independent.