Linear Algebra - Another way of Proving a Basis?
Matthew Harrington 
If we have a Vector Space $V$, can we prove that a set $S$ of vectors $\in V$ is a basis of $V$ given that:
- $S$ contains the same number of vectors as $\dim{(V)}$.
- Every vector in a basis of $V$ can be written as a linear combination of the vectors in S
Example: Let $V$ be $\Bbb R_3$. The Standard Basis of $\Bbb R_3$ is $\{b_1,b_2,b_3\}=\{(1,0,0),(0,1,0),(0,0,1)\}$. Let $S$ be $\{v_1,v_2,v_3\}=\{(1,0,0),(1,1,0),(1,1,1)\}$. Then:
$$ \begin{align} v_1 = b_1 \\ v_2 - v_1 = b_2 \\ v_3 - v_2 = b_3 \end{align} $$ So: $$ \begin{align} c_1b_1+c_2b_2+c_3b_3 = (a,b,c) \\ c_1(v_1)+c_2(v_2-v_1)+c_3(v_3-v_2) = (a,b,c) \\ (c_1-c_2)v_1 + (c_2-c_3)v_2 + c_3v_3 = (a,b,c) \end{align} $$ therefore, since $\{b_1,b_2,b_3\}$ is independent (let $a = b = c = 0$) and spanning, $S$ is also independent and spanning so $S$ is a basis of $V$
If a set $S$ satisfies the before-mentioned conditions, is it a basis?
Edit: in response to Andres Caicedo, yes, $V$ is finite dimensional.
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$\begingroup$Yes, every spanning set contains a basis: you just remove vectors that can be written as a linear combination of the others. So we can remove vectors from $S$ to get a basis. But the resulting basis must have $\dim V$ vectors and that's how many vectors $S$ has. Therefore we removed $0$ vectors to get the basis. The basis is $S$.
Similarly if $|S| = \dim V$ and $S$ is a linearly independent set then $S$ is a basis.
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