Subsets that is Linear independent $A$ 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.
 A: 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.
A: 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.
A: 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 $v\in C.$
