This proof was presented to me:

The set of vectors $B = \{v_1,\ldots,v_n\}$ is linearly independent (is a base for $V$); the only solution of the equation $$\lambda v_1+\ldots+\lambda v_n=0, \lambda_i \in K$$

is $$\lambda_1=\ldots=\lambda_n=0$$ Proof by contradiction: Suppose there is a linearly dependent subset of $m$ vectors, with $m<n$, then there are $\alpha_1,\ldots,\alpha_n$ not all zero, such that: $$\alpha_1 v_1+\ldots+\alpha_n v_n=0, \alpha_i \in K$$ It follows that we could find a solution for $\lambda_{i_k}=\alpha_k$, for $1<k<n$ and for the rest of coefficients, $\lambda_j=0$. Then, if a linearly dependent subset of $B$ exists, $B$ cannot be linearly independent. Therefore, any non-empty subset of $B$ is linearly independent.

  • What's the meaning of $\lambda_{i_k}=\alpha_k$ and $\lambda_j$ in the proof ?
  • Is it necessary to justify why B is linearly independent ?
  • Is this a valid proof ?
  • $\begingroup$ If the set is linearly independent, the coefficients are unique. You can, assume by contradiction one of the vectors can be represented as a linear combination of the rest of the vectors. However, I don't see why an arbitrary linearly independent set should necessary be a basis for $V$ as you say in the beginning. Vectors $v_1,\ldots,v_n$ form a basis for $V\iff \operatorname{span}\{v_1,\ldots,v_n\}=V$. $\lambda_j=0$ if you assume $\{v_1,\ldots,v_k\}\subset\{v_1,\ldots,v_n\}$ is linearly dependent. $\endgroup$ – Invisible Jun 10 '20 at 19:59
  • $\begingroup$ Only a linearly independent set can be extended to a basis for the whole vector space $V$. The point is, linearly independent vectors cannot annihilate each other. If the contradiction confuses you, look at $(\alpha_k-\lambda_{i_k})$. $\endgroup$ – Invisible Jun 10 '20 at 20:05
  • $\begingroup$ Thank you, @Cheesecake ! I was given $B=\{v_1,\ldots,v_n\}$ is a base of $V$. Is it incorrect to conclude $v_1,\ldots,v_n$ are linearly independent ? $\endgroup$ – F. Zer Jun 10 '20 at 20:29
  • 1
    $\begingroup$ If it was given $B$ is a basis, then it is linearly independent. $\endgroup$ – Invisible Jun 10 '20 at 21:24
  • $\begingroup$ I will clarify in the post. $\endgroup$ – F. Zer Jun 10 '20 at 22:17

The proof, as reported, is not really well done, albeit correct.

First, a set $\{v_1,v_2,\dots,v_n\}$ is linearly independent if and only if the only solution to $\lambda_1v_1+\lambda_2v_2+\dots+\lambda_nv_n=0$ is $\lambda_1=\lambda_2=\dots=\lambda_n=0$.

Choose a subset $\{v_{i_1},v_{i_2},\dots,v_{i_k}\}$ of $B$ and suppose $$ \alpha_1v_{i_1}+\alpha_2v_{i_2}+\dots+\alpha_kv_{i_k}=0 \tag{*} $$ For $1\le i_j\le n$ with $1\le j \le k$ , define $$ \lambda_i= \begin{cases} \alpha_{i_j} & \text{if }i=i_j \\[4px] 0 & \text{otherwise} \end{cases} $$ Then clearly $\lambda_1v_1+\lambda_2v_2+\dots+\lambda_nv_n=0$, which implies $\lambda_1=\lambda_2=\dots=\lambda_n=0$ and, in particular, $\alpha_{i_1}=\alpha_{i_2}=\dots=\alpha_{i_k}=0$. QED

Basically, you rewrite the linear combination (*) by inserting the “missing vectors” with zero coefficient. This doesn't change the result of the linear combination (which is zero by assumption), but allows to apply the assumption that the full set is linearly independent.

  • $\begingroup$ Thank you so much, @egreg ! Could you explain what's the meaning of the double indices in $v_{i_k}$? $\endgroup$ – F. Zer Jun 10 '20 at 22:16
  • $\begingroup$ @F.Zer They're just some of the indices among $\{1,2,\dots,n\}$, say $i_1=3,i_2=4,i_3=7$, for example. Another way to do the proof would be: without loss of generality, we can assume that the subset is $\{v_1,v_2,\dots,v_k\}$ (because swapping around the vectors doesn't change the fact that the set is linearly independent). Suppose $\lambda_1v_1+\dots+\lambda_kv_k=0$ and define $\lambda_{k+1}=\dots=\lambda_n=0$; then $\lambda_1v_1+\dots+\lambda_kv_k+\lambda_{k+1}v_{k+1}+\lambda_nv_n=0$ and the thesis follows as before. $\endgroup$ – egreg Jun 10 '20 at 22:18
  • $\begingroup$ So, $i \in \mathbb{Z}$, how can I interpret it ? It is something like $1_1, 3_2$ ? $\endgroup$ – F. Zer Jun 10 '20 at 22:22
  • $\begingroup$ @F.Zer No, it's a customary notation for denoting $k$ integers to be chosen among the given ones. See edit to my previous comment. $\endgroup$ – egreg Jun 10 '20 at 22:23
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    $\begingroup$ @F.Zer The idea is that they're arbitrarily chosen, indeed. $\endgroup$ – egreg Jun 11 '20 at 11:55

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