Let $V$ be a vector space and $ v_1,\dots,v_k \in V$ be a set of linearly independent vectors.

Proof that if $w \in V $ and $ v_1+w,\dots,v_k+w $ linearly dependent, then $ w \in span\{v_1,\dots,v_k\} $.

My solution:


$a_1(v_1+w) + \dots + a_k(v_k+w) = 0$

Because of the linear dependence of $v_1+w,\dots, v_k+w$ there is atleast one $a_i \neq 0 (i=1,\dots,k)$


$a_1v_1+\dots+a_kv_k +xw= 0$ with $x=(a_1+\dots+a_k)$.

Because $v_1, \dots, v_k$ is linearly independent $a_1v_1+\dots+a_kv_k = 0$ has only $a_1=\dots=a_k=0$ as a solution and therefore $x \neq 0$.

So $ w = -\frac{a_1}{x} v_1-\dots-\frac{a_n}{x} v_n$ and $w \in span\{v_1,\dots,v_k\}$.

Is this correct?

  • $\begingroup$ Sorry I read improperly your question. The statement holds but the proof is false. You didn't use the fact that $v_1+w,...,v_k+w$ are linearly dependent (or at least didn't make it clear how you used it to deduce that $x\not = 0$). $\endgroup$ – Yanko Jan 12 at 14:58
  • $\begingroup$ My reasoning why $x \neq0 $ is because of the linearly dependence of $v_1+w, \dots, v_k+w$ $\endgroup$ – strelsol Jan 12 at 15:03
  • $\begingroup$ I understand that. But you need to add two lines, first that you choose $a_1,...,a_k$ where at least one of them is none zero and another line that if by contradiction $x=0$ then $a_1v_1+...+a_kv_k=0$ and so $a_1=...=a_k=0$. $\endgroup$ – Yanko Jan 12 at 15:05
  • $\begingroup$ That is invalid reasoning. You can have linearly dependent vectors $v_1+w,\cdots,v_k+w$ with $a_1(v_1+w)+\cdots+a_k(v_k+w)=0$ and $a_1+\cdots+a_k=0$. For example let $k=2$, $v_1=v_2$, $w=0$, $a_1=1$, $a_2=-1$. $\endgroup$ – Ben W Jan 12 at 15:06
  • $\begingroup$ Your $v_1$ and $v_2$ are linearly dependent $\endgroup$ – strelsol Jan 12 at 15:13

Your proof is sort of correct, but there is an ambiguity here: "$a_1v_1+\cdots+a_kv_k=0$ has only $a_1=\cdots=a_k=0$ as a solution and therefore $x\neq 0$." That is true but it is unclear if you really understand why.

Proof. Let $a_1,\cdots,a_k\in\mathbb{F}$, not all zero, such that $$0=\sum_{i=1}^ka_i(v_i+w).$$ It cannot be that $x:=a_1+\cdots+a_k=0$, otherwise we would have $$0=\sum_{i=1}^ka_i(v_i+w)=\sum_{i=1}^ka_iv_i,$$ contradicting linear independence of $\{v_i\}_{i=1}^k$. Thus $$0=\sum_{i=1}^ka_i(v_i+w)=xw+\sum_{i=1}^ka_iv_i,$$ and hence $$w=-\frac{1}{x}\sum_{i=1}^ka_iv_i\in\text{span}\{v_i\}_{i=1}^k.\;\;\square$$


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