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

  1. 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.
  2. State and prove a generalization of part (a) for the space \mathbb{R}^n.

  1. 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

  2. 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

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