Mathematical Immaturity

1.17 Exercises

• In each case, find an orthonormal basis for the subspace of $V_4$ spanned by the given vectors.
  $\qquad \text{(a)}\quad x_1 = (1, 1, 0, 0), \quad x_2 = (0, 1, 1, 0), \quad x_3 = (0, 0, 1, 1), \quad x_4 = (1, 0, 0, 1).$
  $\qquad \text{(b)}\quad x_1 = (1, 1, 0, 1), \quad x_2 = (1, 0, 2, -1), \quad x_3 = (1, 2, -2, 1).$
• The subspace of $V_4$ spanned by $x_1,$ $x_2,$ $x_3,$ and $x_4$ 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 $x_1,$ $x_2,$ $x_3,$ and $x_4:$ \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 \\ 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 \\ y_4 &= x_4 - \frac{(x_4, y_3)}{(y_3, y_3)}y_3 - \frac{(x_4, y_2)}{(y_2, y_2)}y_2 - \frac{(x_4, y_1)}{(y_1, y_1)}y_1 \end{align*}

$\text{(a)} \quad$ For the subspace of $V_3$ spanned by $x_1 = (1, 1, 0, 0), \quad x_2 = (0, 1, 1, 0), \quad x_3 = (0, 0, 1, 1), \quad x_4 = (1, 0, 0, 1),$ we get: \begin{align*} y_1 &= x_1 = (1, 1, 0, 0) \\ y_2 &= x_2 - \frac{(x_2, y_1)}{(y_1, y_1)}y_1 \\ &= (0, 1, 1, 0) - \frac{1}{2}(1, 1, 0, 0) \\ &= \left(-\frac{1}{2}, \frac{1}{2}, 1, 0 \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 \\ &= (0, 0, 1, 1) - \frac{1}{6/4}\left(-\frac{1}{2}, \frac{1}{2}, 1, 0 \right) - \frac{0}{2}(1, 1, 0, 0) \\ &= \left(\frac{1}{3}, -\frac{1}{3}, \frac{1}{3}, 1\right) \\ y_4 &= x_4 - \frac{(x_4, y_3)}{(y_3, y_3)}y_3 - \frac{(x_4, y_2)}{(y_2, y_2)}y_2 - \frac{(x_4, y_1)}{(y_1, y_1)}y_1 \\ &= (1, 0, 0, 1) - \left(\frac{1}{3}, -\frac{1}{3}, \frac{1}{3}, 1\right) + \frac{1}{3}\left(-\frac{1}{2}, \frac{1}{2}, 1, 0 \right) - \frac{1}{2}(1, 1, 0, 0) \\ &= \left(0, 0, 0, 0\right) \end{align*} Since $y_4 = O,$ the set $\{y_1, y_2, y_3, y_4\}$ is dependent. As such, the orthogonal basis for the subspace of $V_3$ spanned by $x_1 = (1, 1, 0, 0), \ x_2 = (0, 1, 1, 0), \ x_3 = (0, 0, 1, 1), \ x_4 = (1, 0, 0, 1)$ is given by \begin{align*} y_1 &= (1, 1, 0, 0), \quad y_2 = \left(-\frac{1}{2}, \frac{1}{2}, 1, 0 \right), \quad y_3 = \left(\frac{1}{3}, -\frac{1}{3}, \frac{1}{3}, 1\right). \end{align*} With orthonormal basis: \begin{align*} e_1 &= \frac{1}{2}\sqrt{2}(1, 1, 0, 0) \\ e_2 &= \frac{\sqrt{2}}{\sqrt{3}}\left(-\frac{1}{2}, \frac{1}{2}, 1, 0 \right) \\ &= \frac{1}{6}\sqrt{6}(-1, 1, 2, 0) \\ e_3 &= \frac{3}{\sqrt{12}}\left(\frac{1}{3}, -\frac{1}{3}, \frac{1}{3}, 1\right) \\ &= \frac{1}{6}\sqrt{3}(1, -1, 1, 3) \quad \blacksquare \end{align*}

$\text{(b)} \quad$ For the subspace of $V_3$ spanned by $x_1 = (1, 1, 0, 1), \quad x_2 = (1, 0, 2, 1), \quad x_3 = (1, 2, -2, 1),$ we have: \begin{align*} y_1 &= x_1 = (1, 1, 0, 1) \\ y_2 &= x_2 - \frac{(x_2, y_1)}{(y_1, y_1)}y_1 \\ &= (1, 0, 2, 1) - \frac{2}{3}(1, 1, 0, 1) \\ &= \frac{1}{3}\left(1, -2, 6, 1\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 \\ &= (1, 2, -2, 1) - \frac{-14/9}{42/9}(1, -2, 6, 1) - \frac{4}{3}(1, 1, 0, 1) \\ &= \frac{1}{3}(3, 6, -6, 3) + \frac{1}{3}(1, -2, 6, 1) - \frac{1}{3}(4, 4, 0, 4) \\ &= (0, 0, 0, 0) \end{align*} Since $y_3 = O,$ the orthogonal basis for the subspace spanned by $$x_1 = (1, 1, 0, 1), \quad x_2 = (1, 0, 2, 1), \quad x_3 = (1, 2, -2, 1)$$ is given by $y_1 = (1, 1, 0, 1), y_2 = \frac{1}{3}(1, -2, 6, 1).$ Because $y_1$ and $y_2$ are orthogonal, we can multiply $y_2$ by $3$ to simplify the fractional scalar and yield the orthonormal basis: \begin{align*} e_1 &= \frac{1}{3}\sqrt{3}(1, 1, 0, 1), \quad e_2 = \frac{1}{\sqrt{42}}(1, -2, 6, 1) \quad \blacksquare \end{align*}