# Linear Algebra - Another way of Proving a Basis?

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:

1. $S$ contains the same number of vectors as $\dim{(V)}$.
2. 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.

• What is then dimension of the vector space? – Artem May 16 '13 at 21:51
• I got lost: what is the question? – DonAntonio May 16 '13 at 21:52
• @HenryT.Horton I think the OP means $S$ satisfies both 1 and 2. In this case, the reasoning looks correct. – Ayman Hourieh May 16 '13 at 21:52
• yes! That's because such a set has to be linearly independent or else you'd get a contradiction to $|S|=$dim $V$ – user64480 May 16 '13 at 21:53
• Note that this only works if $V$ is finite dimensional. Probably this is the case you had in mind, so you may want to add this assumption. Otherwise, the answer is no. – Andrés E. Caicedo May 16 '13 at 22:23

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.