# Linear independence of set with added vector

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:

Suppose

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

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

Then

$$a_1v_1+\dots+a_kv_k +xw= 0$$ 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?

• 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 $x\not = 0$). – Yanko Jan 12 at 14:58
• My reasoning why $x \neq0$ is because of the linearly dependence of $v_1+w, \dots, v_k+w$ – strelsol Jan 12 at 15:03
• 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 $a_1=...=a_k=0$. – Yanko Jan 12 at 15:05
• 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 $k=2$, $v_1=v_2$, $w=0$, $a_1=1$, $a_2=-1$. – Ben W Jan 12 at 15:06
• Your $v_1$ and $v_2$ are linearly dependent – 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 $$w=-\frac{1}{x}\sum_{i=1}^ka_iv_i\in\text{span}\{v_i\}_{i=1}^k.\;\;\square$$