Mathematical Immaturity

1.13 Exercises

• In a complex Euclidean space, prove that the inner product has the following properties for all elements $x, y, z,$ and all complex $a$ and $b.$
  $\text{(a)}\quad (ax, by) = a\overline{b}(x, y).$
  $\text{(b)}\quad (x, ay + bz) = \overline{a}(x, y) + \overline{b}(x, z).$

• $\text{(a)} \quad$ From the modified commutativity axiom described in Section 11, we know that $(ax, by) = \overline{(by, ax)},$ where $\overline{(by, ax)}$ denotes the complex conjugate of $(by, ax).$ Then, by associativity, we have $\overline{(by, ax)} = \bar{b}\,\overline{(y, ax)}.$ Applying Axiom (1') and associativity once more, we get \begin{align*} \bar{b}\,\overline{(y, ax)} &= \overline{b}(ax, y) \\ &= a\bar{b}(x, y). \quad \blacksquare \end{align*}

  $\text{(b)} \quad$ By distributivity, we have $$(x, ay + bz) = (x, ay) + (x, bz)$$ Then, applying Axiom (1') and associativity to the right-hand side, we get \begin{align*} (x, ay) &= \overline{(ay, x)} \\ &= \bar{a}\overline{(y, x)} \\ &= \bar{a}(x, y) \\ \\ (x, bz) &= \overline{(bz, x)} \\ &= \bar{b}\overline{(z, x)} \\ &= \bar{b}(x, z) \end{align*} Then, we can express their sum as: $$(x, ay + bz) = \bar{a}(x, y) + \bar{b}(x, z) \quad \blacksquare$$