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Let $B = \{v_1, v_2, ..., v_k\} \in \mathbb{R}^n$ and $B$ is linearly independent. Prove that if $\overrightarrow{v} \not \in \text{span } B$ then $\{v_1, v_2, ..., v_k, v\}$ is linearly Independent.

Proof:

Assume $\{v_1, v_2, ..., v_k, v \}$ is linearly dependent. Then because $\{v_1, v_2, ..., v_k\}$ is linearly independent from hypothesis, it must follow that $v \in \text{span } B$, contradiction.

Is this correct? If not, can you tell me why?

Edit:

We have $c_1v_1 + c_2v_2 + ... + c_kv_k = 0$ for $c_1 = c_2 = ... = c_k= 0 $.Since $B + v$ is linearly dependent, we must have

$c_1v_1 + c_2v_2 + ... + c_kv_k + xv = 0$ where $x \ne 0$, thus it follows that

$-c_1v_1/x - c_2v_2/x - ... - c_kv_k/x = v$ thus $v \in \text{span } B$

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  • $\begingroup$ It's not wrong, but I think you're a bit brief. I would personally prefer if you expand a bit on why this means $v\in \operatorname{Span} B$. Something with an explicit linear combination. $\endgroup$ – Arthur Jan 24 '17 at 6:59
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    $\begingroup$ @Arthur, how is it with the Edit? $\endgroup$ – Aditya Kalra Jan 24 '17 at 7:05
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If $\{v_1, v_2, ..., v_k, v\}$ is linearly dependent there exist scalars $\lambda_i$ not all null such that $$\lambda_1v_1+\lambda_2 v_2+\cdots \lambda_k v_k+ \lambda_{k+1}v=0$$ As $\{v_1, v_2, ..., v_k\}$ is linearly independent, necessarily $\lambda_{k+1}\ne0.$ So, $$v=\left(-\frac{\lambda_1}{\lambda_{k+1}}\right)v_1+\left(-\frac{\lambda_2}{\lambda_{k+1}}\right)v_2+\cdots +\left(-\frac{\lambda_k}{\lambda_{k+1}}\right)v_k\in \text{Span }B\text{ (contradiction).}$$

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  • $\begingroup$ I'm probably missing something here, but what exactly is being contradicted and how does it prove anything about the original problem statement? $\endgroup$ – infinitylord Jan 24 '17 at 7:42
  • $\begingroup$ $$v \not \in \text{Span } B\Rightarrow\{v_1, v_2, ..., v_k, v\}\text{ is linearly independent}$$ is equivalent to $$\{v_1, v_2, ..., v_k, v\}\text{ is linearly dependent}\Rightarrow v \in \text{Span } B.$$ $\endgroup$ – Fernando Revilla Jan 24 '17 at 7:50
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It's much better with the edit. While all the elements are there, I would perhaps phrase that proof a bit differently: start with $c_1v_1 + c_2v_2 + \ldots + c_kv_k + xv = 0$ where not all coefficients are $0$. Argue that $x\neq0$ (since otherwise we would have $c_1v_1 + c_2v_2 + \ldots + c_kv_k = 0$, which implies that all coefficients are $0$), which means that $-c_1v_1/x - c_2v_2/x - \ldots - c_kv_k/x = v$ makes sense.

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