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?

  • $\begingroup$ sorry, linearly independent from what? Or do you mean a basis of $n$ vectors $v_1,...,v_n$? $\endgroup$ Commented May 7, 2014 at 17:10
  • $\begingroup$ I guess it follows. $\endgroup$
    – MoDzeus
    Commented May 7, 2014 at 17:11
  • $\begingroup$ @Modestas_S are you assuming that the $u_i$ are also linearly independent? $\endgroup$
    – essay
    Commented May 7, 2014 at 17:15

2 Answers 2


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 $x+y=(0,\dots,0)$


In general the answer is no.

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.

Hope this was clear.


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