Prove that a set is linearly independent in R3 iff it contains the three unit coordinate vectors

Prove that a set $S$ of three vectors in $\mathbb{R}^3$ is linearly independent if and only if its span contains the three unit coordinate vectors.
State and prove a generalization of part (a) for the space $\mathbb{R}^n$ .

Proof. Assume $S$ is a basis for $\amthbb{R}^3$ . Then $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are in the span of $S$ since they are in $\mathbb{R}^3$ and the span of $S$ is all of $\mathbb{R}^3$ by the definition of basis.
Conversely, assume $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are in the span of $S$ . Let $S = \{ A,B,C \}$ . Since $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are in the span of $S$ we know there are scalars $a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3$ such that

$\begin{align*} \mathbf{i} &= a_1 A + b_1 B + c_1 C \\ \mathbf{j} &= a_2 A + b_2 B + c_2 C \\ \mathbf{k} &= a_3 A + b_3 B + c_3 C. \end{align*}$

But then if $(x,y,z) \in \mathbb{R}^3$ is any vector we have

$\begin{align*} (x,y,z) &= x \mathbf{i} + y\mathbf{j} + z \mathbf{k} \\ &= x (a_1 A + b_1 B + c_1 C) + y(a_2 A + b_2 B + c_2 C) + z (a_3 A + b_3 B + c_3 C) \\ &= (xa_1 +ya_2 + za_3)A + (xb_1 + yb_2 + zb_3)B + (xc_1 + yc_2 + zc_2) C. \end{align*}$

Then, if $L(S) = \mathbb{R}^3$ for three vectors in $\mathbb{R}^3$ then $S$ is a basis (the three vectors must be independent, otherwise there would be two vectors that span $\mathbb{R}^3$ , contradicting Theorem $12.10.) \qquad \blacksquare$
Claim. A set $S$ of $n$ vectors in $\mathbb{R}^n$ is a basis for $\mathbb{R}^n$ if and only if its linear span $L(S)$ contains the unit coordinate vectors of $\mathbb{R}^n$ .
Proof. Assume $S$ is a basis for $\mathbb{R}^n$ , then $L(S) = \mathbb{R}^n$ . Since the unit coordinate vectors of $\mathbb{R}^n$ are in $\mathbb{R}^n$ they are then in $L(S)$ .
Conversely, assume the unit coordinate vectors are in $L(S)$ . Then, by the same argument as in part (a) we know $\mathbb{R}^n \subseteq L(S)$ . But, since there are $n$ vectors in $S$ , we know $L(S) \subseteq \mathbb{R}^n$ . Hence, $L(S) = \mathbb{R}^n$ . Furthermore, the $n$ vectors of $S$ are independent since they span $\mathbb{R}^n$ (if they were dependent, then there would be a set $S'$ with fewer than $n$ vectors which would span $\mathbb{R}^n$ , contradicting Theorem $12.10.) \qquad \blacksquare$