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Let $S = \{v_1, .., v_k\} $ be a set in $\mathbb{R}^n$ where $v_1, .. v_k$ are vectors. Prove that if $\text{span } S = \mathbb{R}^n$ then $k \ge n$.

We are essentially doing $c_1v_1 + .. + c_kv_k = b$, where $b$ is a vector in $\mathbb{R}^n$ and $c_1, .. , c_k \in \mathbb{R}$.

By one of the theorems, it follows that $\text{rank } A = n$, where $A$ is the coefficient matrix $[v_1, v_2, .. v_k]$

I know the system contains $k - \text{rank }A = k - n$ free variables

But there must be $0$ free variables, then isn't $k = n$?

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    The linear span of $k$ vectors has *dimension* at most $k$ while $\mathbb{R}^n$ has dimension $n$. Here, you can think of dimension as the smallest cardinality of a family spanning a linear space. – 2017-02-05
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    @heptagon, not allowed to use dimension – 2017-02-05
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    Why do you think there must be 0 free variables? – 2017-02-05
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    @i707107, each vector has a unique representation in the span doesnt it? – 2017-02-05
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    @Amad27 That is not true in general. Only true if $k=n$. – 2017-02-05
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    Consider the three vectors $x=(0,1)$, $y=(1,0)$, and $z=(1,1)$ in $\Bbb R^2$; then it's not true that every vector in the span of $\{x,y,z\}$ has a unique representation (indeed, no vector does!). For example, $1x+1y+0z=0x+0y+1z$. – 2017-02-05

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