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Let $S=\{u_1,u_2,\cdots,u_n\}$ be a spanning set for an n-dimensional vector space $V$. Show $S$ is linearly independent.

My try:

Since $S$ is spanning set for $V$, any vector $x$ in $V$ can be written as the linear combination of vectors in $S$ as

$$ x = \alpha_1 u_1 + \cdots + \alpha_n u_n $$

Since $V$ is a vector space, $0 \in V$. Let $x=0$.

$$ 0= \alpha_1 u_1 + \cdots + \alpha_n u_n $$

We need to show $\alpha_i$'s are zero.

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    $\begingroup$ @poyea I think the OP most likely meant to prove this statement under the condition that $dim(V) = n$. $\endgroup$ – user82261 Mar 11 '19 at 4:31
  • $\begingroup$ Oh thanks @user82261 $\endgroup$ – poyea Mar 11 '19 at 4:34
  • $\begingroup$ Recall that if $dim(V) = n$, then any set containing fewer than $n$ elements can't be a spanning set. Now use the answer below. This result is implicitly used in the answer below to hint at the desired contradiction. Ultimately, you'll have to use the relationship that in a finite dimensional vector space of dimension $n$, a set is a spanning set and a linearly independent set if and only if it contains $n$ elements, no more and no less. $\endgroup$ – user82261 Mar 11 '19 at 4:34
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Basically, if $\alpha_j\neq 0,$ then $u_j$ is in the span of the other $n-1$ vectors, which means $Span(S)$ could not have been $n$ dimensional in the first place. You can write this in more detail and get yourself a proof by contradiction.

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  • $\begingroup$ Do you mean from the beginning I should assume what you assumed without writing any other things? $\endgroup$ – Sepide Mar 11 '19 at 5:27
  • $\begingroup$ @Sepide you can say "let's prove by contradiction" or "Suppose to the contrary" etc. Or how you never have done a proof before? $\endgroup$ – dezdichado Mar 11 '19 at 14:56
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Proving the contrapositive is not easier in my opinion, but it does give you another way to think about the problem.

$$ \text{$S$ spans $V$} \implies \text{$S$ is linearly independent} \tag{Original} $$

is equivalent to

$$ \text{$S$ is linearly dependent} \implies \text{$S$ does not span $V$} \tag{Contrapositive}$$

Let's march through $u_1, u_2, \dots u_n$ and keep $u_i$ if it is linearly independent of the preceding vectors ($u_1 \dots u_{i-1}$). Let's call the new collection of linearly independent vectors $w_1, \dots, w_m$ where $m < n$. We know $m$ is less than $n$ because at least one $u_i$ must be excluded. We know that $w_1, \dots, w_m$ and $u_1, \dots, u_n$ span the same subspace because we only excluded $u_i$s that were in the "span so far".

$w_1, \dots, w_m$ spans a subspace of exactly dimension $m$, but $m < n$ by our hypothesis.

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