Mathematical Immaturity

1.17 Exercises

• In each case, find an orthonormal basis for the subspace of $V_3$ spanned by the given vectors.
  $\qquad \text{(a)}\quad x_1 = (1, 1, 1), \quad x_2 = (1, 0, 1), \quad x_3 = (3, 2, 3).$
  $\qquad \text{(b)}\quad x_1 = (1, 1, 1), \quad x_2 = (-1, -1, -1), \quad x_3 = (1, 0, 1).$
• The subspace of $V_3$ spanned by $x_1,$ $x_2,$ and $x_3$ becomesa a real Euclidean space with the inner product defined as the dot product: $$(x, y) = x \cdot y.$$ With this, we can use the Gram-Schmidt process to construct an orthogonal basis for the subspace spanned by $x_1,$ $x_2,$ and $x_3:$ \begin{align*} (1.16) \qquad y_1 = x_1, \quad y_{r+1} &= x_{r+1} - \sum_{i=1}^r \frac{(x_{r+1}, y_i)}{(y_i, y_i)}y_i \quad \text{for} \quad r = 1, 2, \dots, k-1 \end{align*} Giving us: \begin{align*} y_1 &= x_1 = (1, 1, 1) \\ y_2 &= x_2 - \frac{(x_2, y_1)}{(y_1, y_1)}y_1 \\ y_3 &= x_3 - \frac{(x_3, y_2)}{(y_2, y_2)}y_2 - \frac{(x_3, y_1)}{(y_1, y_1)}y_1 \end{align*}

  $\text{(a)}\quad$ For $x_1 = (1, 1, 1), \quad x_2 = (1, 0, 1), \quad x_3 = (3, 2, 3),$ we get: \begin{align*} y_1 &= x_1 = (1, 1, 1) \\ y_2 &= x_2 - \frac{(x_2, y_1)}{(y_1, y_1)}y_1 \\ &= (1, 0, 1) - \frac{2}{3}(1, 1, 1) \\ &= \left(\frac{1}{3}, -\frac{2}{3}, \frac{1}{3}\right) \\ y_3 &= x_3 - \frac{(x_3, y_2)}{(y_2, y_2)}y_2 - \frac{(x_3, y_1)}{(y_1, y_1)}y_1 \\ &= (3, 2, 3) - \frac{2/3}{6/9} \left(\frac{1}{3}, -\frac{2}{3}, \frac{1}{3}\right) - \frac{8}{3}(1, 1, 1) \\ &= \frac{1}{3}\left[(3, 2, 3) - (1, -2, 1) - (8, 8, 8)\right] \\ &= \frac{1}{3}(-6, 4, -6) \end{align*} As we can see, $y_3 = -2y_2,$ making the set $\{y_1, y_2, y_3\}$ dependent. Thus, the orthogonal basis for the subspace of $V_3$ spanned by $x_1 = (1, 1, 1), \quad x_2 = (1, 0, 1), \quad x_3 = (3, 2, 3)$ is \begin{align*} y_1 &= (1, 1, 1), \quad y_2 = \frac{1}{3}(1, -2, 1) \end{align*} Dividing these vectors by their respective norms, we get the orthonormal basis \begin{align*} e_1 &= \frac{1}{3}\sqrt{3}(1, 1, 1), \quad e_2 = \frac{1}{6}\sqrt{6}(1, -2, 1) \quad \blacksquare \end{align*}

  $\text{(b)} \quad$ For $x_1 = (1, 1, 1), \quad x_2 = (-1, -1, -1), \quad x_3 = (1, 0, 1),$ we first note that $x_2$ is a scalar multiple of $x_1.$ As such, the subspace spanned by $x_1,$ $x_2,$ and $x_3$ is the same as the one spanned by $x_1$ and $x_3.$ Accordingly, we use the Gram-Schmidt process on $x_1$ and $x_3$ to find the orthogonal basis that spans this subspace: \begin{align*} y_1 &= x_1 = (1, 1, 1) \\ y_2 &= x_3 - \frac{(x_3, y_1)}{(y_1, y_1)}y_1 \\ &= (1, 0, 1) - \frac{2}{3}(1, 1, 1) \\ &= \frac{1}{3}(1, -2, 1) \end{align*} As we can see, these two vectors form the same orthogonal basis as in part (a), giving us the same orthonormal basis: \begin{align*} e_1 &= \frac{1}{3}\sqrt{3}(1, 1, 1), \quad e_2 = \frac{1}{6}\sqrt{6}(1, -2, 1) \quad \blacksquare \end{align*}