# Prove that if $S\equiv\{v_1,…,v_n\}$ is linearly independent, then every subset of $S$ is also linearly independent

I need help with the proof of this statement:

If $S=\big\{ \vec{v_1},...,\vec{v_{n}} \big\}$ is a set of linearly independent vectors then any non-empty subset of $S$ contains only linearly independent vectors.

I tried using the definition

$\vec{v_1},...,\vec{v_n}$ are l.i. $\iff \lambda_1\vec{v_1}+...+\lambda_n\vec{v_n}=\vec{o} \implies \lambda_1=...=\lambda_n=0$

But is this enough to conclude that, for example, considering a subset $S'=\big\{ \vec{v_1},...,\vec{v_{n-1}} \big\}$ for sure $\lambda_1\vec{v_1}+...+\lambda_n\vec{v_{n-1}}=\vec{o} \implies \lambda_1=...=\lambda_{n-1}=0$ ?

It makes sense of course but is it enough for the proof?

• The thing is, if $S'=\{\vec{v_1},\ldots,\vec{v_i}\}$ contained linearly dependent vectors, then you could choose $\lambda_1,\ldots,\lambda_i$ not all $0$ such that the linear combination $\lambda_1\vec{v_1}+\cdots+\lambda_i\vec{v_i}$ is $\vec{0}$. But then $\vec{v_1},\ldots,\vec{v_n}$ would not be linearly independent since you could take those $\lambda_1,\ldots,\lambda_i$ and set the other $\lambda$ to $0$ to get the $\vec{0}$ vector as a linear combination of $\vec{v_1},\ldots,\vec{v_n}$ with not all the $\lambda$ equal to $0$. – Guest Dec 12 '15 at 18:00
• It is enough :) – Babai Dec 12 '15 at 18:23

There is a theorem that states

If a set of vectors $S$ is linearly independent, then any subset of $S$ is also a linearly independent set of vectors.

Proof

Let T be a subset of S. Let us reorder the vectors of S in a way to help us with the proof.

Make $T = \big \{ v_1, \dots ,v_k \big \}$ and $S = \big \{ v_1, \dots ,v_k, \dots , v_n \big \}$

Let's look at the equation $$c_1v_1 + c_2v_2 + \dots + c_kv_k = 0$$

Make $c_{k+1} =c_{k+2} = \dots = c_n = 0$ then we can rewrite the initial equation as $$c_1v_1 + c_2v_2 + \dots + c_kv_k + 0v_{k+1} + \dots + 0v_{n} = 0$$

Since S is linearly independent, then this forces $c_1 = c_2 = \dots = ck = 0$.

$\implies$ T is linearly independent.