# Linearly independent subset of V ⊆ basis of V

The Theorem states:

Let V be a vector space that has a finite spanning set, and let S be a linearly independent subset of V. Then there exists a basis S' of V, with S ⊆ S'

I don't need a proof necessarily, just want to build an intuition for it. Also, how would I use this theorem to solve questions like this one:

a) Find a basis of R^4 containing the linearly independent set S = {(1,2,3,4),(-1,0,0,0)}.

• A linearly independent set uniquely describes the vectors within its span. The theorem says that the unique description that was assigned previously by the linearly independent set doesn't have to be "rewritten" to describe any other vector in the space. – Robert Wolfe Jan 31 '18 at 18:49
• That theorem is of the upmost importance. It simply says that any linearly independent set can be enlarged to a basis. – DonAntonio Jan 31 '18 at 19:05

The idea is that $S$ is a basis for some subspace $W_1\subset V$, so if we choose any $v_1\in in V\backslash W_1$, then $S\cup\{v_1\}$ is linearly independent, and spans a subspace $W_2$ with $W_1\subset W_2\subseteq V$. If $W_2=V$ we are done, otherwise, we can find $v_2\in V\backslash W_2$ so that $S\cup\{v_1,v_2\}$ is linearly independent. Keep going until you have constructed a basis.
In your particular example, find a vector that is not in the span of $S$ (e.g. $(0,1,0,0)$ is an easy choice). Then $S'=\{(1,2,3,4),(-1,0,0,0),(0,1,0,0)\}$ is linearly independent and spans a $3$-dimensional subspace. Find a vector not in the span of $S'$ to form a set $S''$ that is a basis.
Take a basis $\{v_1,v_2,v_3,v_4\}$ of $\mathbb{R}^4$ (the standard basis, for instance). Consider the set $S_1=S\cup\{v_1\}$. Is it linearly independent? If it is, keept it; otherwise, call $S_1$ to the set $S\cup\{v_2\}$ and so on. At a certain point, you'll have a linearly independent set $S_1$ with $3$ elements. Now, you start all over again (except that there's no need to try again an element of $\{v_1,v_2,v_3,v_4\}$ which has already been tested). So, suppose that $S_1=S\cup\{v_2\}$ (and this will be true if $\{v_1,v_2,v_3,v_4\}$ is the standard basis). Now, you start all over again, doing $S_2=S_1\cup\{v_3\}$. When you get a set with $4$ linearly independent vectors, that will be your basis.