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$A$ is the set of linearly independent vectors and $B$ is a nonempty subset of $A$. Then is $B$ also linearly independent?

I know that this is true, since $A$ is LI and this means that no vector in $A$ is a linear combination of others. Then since $B$ is a subset, $B$ is also linear independent.

But I do not know how to show this.

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    $\begingroup$ You just did show it. Now write it more formal. If $(v_1,…,v_n)$ linear independant, then $$\sumλ_iv_i =0 ⇒ λ_i=0, i=1,…,n.$$ Let $w_1,…,w_m$ be a subset ... $\endgroup$ – P. Siehr Nov 13 '17 at 13:54
  • $\begingroup$ I have to disagree, you didn't already show it. You just wrote the definition of linear independence and then stated the result you want to proof. Admittedly, the result follows easily from the definition, but you still need to write out, why exactly $B$ being a subset of $A$ implies linear independence. $\endgroup$ – klirk Nov 18 '17 at 23:28
  • $\begingroup$ Two ways to prove it would be either: 1) Suppose $v \in B$. As $B \subset A$ and $A$ is linearly independent, it follows that... 2)Proof by contradiction: Assume for contradiction, that $B$ is not linear independent, then ... $\endgroup$ – klirk Nov 18 '17 at 23:31
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We know that $A$ is a linearly independent set of vectors and $ B \subseteq A$. The set $A$ being linearly independent means that, if $x_1,x_2,...,x_n \in A$ and $\alpha_1, \alpha_2,...,\alpha_n$ are scalars such that $$\alpha_1 x_1 +\alpha_2 x_2 + ...+\alpha_nx_n = 0$$ then we have that $\alpha_1 = \alpha_2 = \ ...\ = \alpha_n = 0$.

Now let $B =\{b_1, b_2, ... , b_k \} \subseteq A$, and let's fix a linear combination of vectors from $B$ such that $$\alpha_1 b_1 +\alpha_2 b_2 + ...+\alpha_kb_k = 0$$ Since $B \subseteq A$, this is, $\{b_1, b_2, ... , b_k \} \subseteq A$ and since $A$ is linearly independent, then this means that $\alpha_1 = \alpha_2 = ... = \alpha_k = 0$, which makes $B$ linearly independent.

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Assume that $A$ is linearly independent and $B \subseteq A$.

Take any $v_1, \ldots, v_n \in B$ and scalars $\lambda_1, \ldots, \lambda_n$ such that

$$\sum_{i=1}^n \lambda_iv_i = 0$$

Now, since $B\subseteq A$, we have that also $v_1, \ldots, v_n \in A$. So, $\sum_{i=1}^n \lambda_iv_i$ is a linear combination of vectors in $A$, equal to $0$. Because $A$ is linearly independent, we obtain that the scalars are $0$:

$$\lambda_1 = \ldots = \lambda_n = 0$$

Since $v_1, \ldots, v_n$ were arbitrary, this proves that $B$ is linearly independent.

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A linearly independent set of vectors need not be a finite set. The definition of "$A$ is linearly independent" is that if $C$ is any non-empty finite subset of $A$ and $\sum_{v\in C}vk_v=0$ then $k_v=0$ for all $v\in C.$

Suppose $A$ is linearly independent and $B\subset A.$ Then any finite non-empty $C\subset B$ is a finite non-empty subset of $A$, so if $\sum_{v\in C}vk_v=0$ then $k_v=0$ for all $v\in C.$

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