Mathematical Immaturity

1.13 Exercises

• Let $x = (x_1, \ldots, x_n)$ and $y = (y_1, \ldots, y_n)$ be arbitrary vectors in $V_n$. In each case, determine whether $(x, y)$ is an inner product for $V_n$ if $(x, y)$ is defined by the formula given. In case $(x, y)$ is not an inner product, tell which axioms are not satisfied.
  $\text{(a)}\quad$ $(x, y) = \sum_{i=1}^n x_i |y_i|$.
  $\text{(b)}\quad$ $(x, y) = \left| \sum_{i=1}^n x_i y_i \right|$.
  $\text{(c)}\quad$ $(x, y) = \sum_{i=1}^n x_i \sum_{j=1}^n y_j$.
  $\text{(d)}\quad$ $(x, y) = \left( \sum_{i=1}^n x_i^2 y_i^2 \right)^{1/2}$.
  $\text{(e)}\quad$ $(x, y) = \sum_{i=1}^n (x_i + y_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2$.
• Recall the definition of the inner product from Section 11:
  
  Definition. $\quad$ A real linear space $V$ is said to have an inner product if for each pair of elements $x$ and $y$ in $V$ there corresponds a unique real number $(x, y)$ satisfying the following axioms for all choices of $x,$ $y,$ and $z$ in $V$ and all real scalars $c.$
  $\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry).
  $\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  $\text{(3)}\quad$ $c(x, y) = (cx, y)\qquad$ (associativity, or homogeneity).
  $\text{(4)}\quad$ $(x, x) > 0\quad$ if $x \neq O \qquad$ (positivity).

• $\text{(a)}\quad$ $(x, y) = \sum_{i=1}^n x_i |y_i|$.
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry).
  \begin{align*} (y, x) &= \sum_{i=1}^n y_i|x_i| \\ &= \sum_{i=1}^n |x_i|y_i \\ &\neq \sum_{i=1}^n x_i|y_i| \\ \\ (y, x) &\neq (x, y). \end{align*} $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  \begin{align*} (x, y + z) &= \sum_{i=1}^n x_i|y_i + z_i| \\ \\ (x, y) + (x, z) &= \sum_{i=1}^n x_i|y_i| + \sum_{i=1}^n x_i|z_i| \\ &= \sum_{i=1}^n x_i(|y_i| + |z_i|) \\ &\neq (x, y+z) \end{align*} $\quad\text{(3)}\quad$ $c(x, y) = (cx, y) \qquad$ (associativity, or homogeneity). \begin{align*} c(x, y) &= c\sum_{i=1}^n x_i|y_i| \\ &= \sum_{i=1}^n cx_i|y_i| \\ &= (cx, y) \end{align*} $\quad\text{(4)}\quad$ $(x, x) > 0 \quad$ if $\quad x \neq O\qquad$ (positivity).
  $\quad$Let $x = (x_1, \dots, x_n)$ be a vector in $V_n$ such that each $x_i \lt 0$ for $i = 1, \cdots, n.$ Then we have: \begin{align*} (x, x) &= \sum_{i = 1}^n x_i|x_i| \\ &= \sum_{i = 1}^n x_i|x_i| \\ &= -\sum_{i = 1}^n x_i^2 \\ &\lt 0 \end{align*} As we can see, $(x, y) = \sum_{i=1}^n x_i|y_i|$ is not an inner product since it violates axioms $(1),$ $(2),$ and $(4). \quad \blacksquare$

  $\text{(b)}\quad$ $(x, y) = \left|\sum_{i=1}^n x_i y_i\right|$.
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (y, x) &= \left|\sum_{i=1}^n y_ix_i\right| \\ &= \left|\sum_{i=1}^n x_iy_i\right| \\ &= (x, y). \end{align*} $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  \begin{align*} (x, y + z) &= \left|\sum_{i = 1}^n x_i(y_i + z_i)\right| \\ &= \left|\sum_{i = 1}^n x_iy_i + \sum_{i = 1}^n x_iz_i\right| \\ &= \left|\sum_{i = 1}^n x_i(y_i + z_i)\right| \\ \\ (x, y) + (x, z) &= \left|\sum_{i = 1}^n x_iy_i\right| + \left|\sum_{i = 1}^n x_iz_i\right| \\ &\neq (x, y + z). \end{align*} $\quad\text{(3)}\quad$ $c(x, y) = (cx, y) \qquad$ (associativity, or homogeneity). \begin{align*} c(x, y) &= c\left|\sum_{i=1}^n x_iy_i\right| \\ \\ (cx, y) &= \left|\sum_{i=1}^n cx_iy_i\right| \\ &= |c|\left|\sum_{i=1}^n x_iy_i\right| \\ &\neq c(x, y) \end{align*} $\quad\text{(4)}\quad$ $(x, x) > 0 \quad$ if $\quad x \neq O\qquad$ (positivity).
  \begin{align*} (x, x) &= \left|\sum_{i = 1}^n x_i^2\right| \\ &\gt 0 \quad \text{if} \quad x \neq O. \end{align*} As we can see, $(x, y) = \left|\sum_{i=1}^n x_iy_i\right|$ is not an inner product since it violates axioms $(2)$ and $(3). \quad \blacksquare$

  $\text{(c)}\quad$ $(x, y) = \sum_{i=1}^n x_i + \sum_{j=1}^n y_j.$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (x, y) &= \sum_{i=1}^n x_i + \sum_{j=1}^n y_j \\ &= \sum_{i=1}^n y_i + \sum_{j=1}^n x_j \\ &= (y, x) \end{align*} $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  \begin{align*} (x, y+z) &= \sum_{i=1}^n x_i + \sum_{j=1}^n y_j + z_j \\ \\ (x, y) + (x, z) &= \left(\sum_{i=1}^n x_i + \sum_{j=1}^n y_j\right) + \left(\sum_{i=1}^n x_i + \sum_{j=1}^n z_j\right) \\ &= 2\sum_{i=1}^n x_i + \sum_{j=1}^n y_j + z_j \\ &\neq (x, y+z) \end{align*} $\quad\text{(3)}\quad$ $c(x, y) = (cx, y) \qquad$ (associativity, or homogeneity). \begin{align*} c(x, y) &= c\left(\sum_{i=1}^n x_i + \sum_{j=1}^n y_j\right) \\ &= c\sum_{i=1}^n x_i + c\sum_{j=1}^n y_j \\ \\ (cx, y) &= c\sum_{i=1}^n x_i + \sum_{j=1}^n y_j \\ &\neq c(x, y) \end{align*} $\quad\text{(4)}\quad$ $(x, x) > 0 \quad$ if $\quad x \neq O\qquad$ (positivity).
  $\quad$Let $x_i \lt 0$ for $i = 1, \dots, n.$ Then we have: \begin{align*} (x, x) &= \sum_{i=1}^n x_i + \sum_{i=1}^n x_i \\ &= 2\sum_{i=1}^n x_i \\ &\lt 0 \end{align*} As we can see, $(x, y) = \sum_{i=1}^n x_i + \sum_{j=1}^n y_j$ is not an inner product since it violates axioms $(2),$ $(3),$ and $(4). \quad \blacksquare$

  $\text{(d)}\quad$ $(x, y) = \left(\sum_{i=1}^n x_i^2y_i^2\right)^{1/2}.$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (x, y) &= \left(\sum_{i=1}^n x_i^2y_i^2\right)^{1/2} \\ &= \left(\sum_{i=1}^n y_i^2x_i^2\right)^{1/2} \\ &= (y, x) \end{align*} $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  \begin{align*} (x, y + z) &= \left[\sum_{i=1}^n x_i^2(y_i + z_i)^2\right]^{1/2} \\ &= \left[\sum_{i=1}^n x_i^2(y_i^2 + 2y_iz_i + z_i^2)\right]^{1/2} \\ \\ (x, y) + (x, z) &= \left(\sum_{i=1}^n x_i^2y_i^2\right)^{1/2} + \left(\sum_{i=1}^n x_i^2z_i^2\right)^{1/2} \\ \\ &\neq (x, y+z) \end{align*} $\quad\text{(3)}\quad$ $c(x, y) = (cx, y) \qquad$ (associativity, or homogeneity). \begin{align*} c(x, y) &= c\left(\sum_{i=1}^n x_i^2y_i^2\right)^{1/2} \\ \\ (cx, y) &= \left(\sum_{i=1}^n c^2x_i^2y_i^2\right)^{1/2} \\ &= |c|\left(\sum_{i=1}^n x_i^2y_i^2\right)^{1/2} \end{align*} $\quad$If $c \lt 0,$ then $c(x, y) \neq (cx, y).$
  $\quad\text{(4)}\quad$ $(x, x) > 0 \quad$ if $\quad x \neq O\qquad$ (positivity).
  \begin{align*} (x, x) &= \left(\sum_{i=1}^n x_i^2x_i^2\right)^{1/2} \\ &= \left(\sum_{i=1}^n x_i^4\right)^{1/2} \\ &\gt 0 \quad \text{if} \quad x \neq O. \end{align*} As we can see, $(x, y) = \left(\sum_{i=1}^n x_i^2y_i^2\right)^{1/2}$ is not an inner product since it violates axioms $(2)$ and $(3). \quad \blacksquare$

  $\text{(e)}\quad$ $(x, y) = \sum_{i=1}^n (x_i + y_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2.$
  $\quad\text{(1)}\quad$ $(x, y) = (y, x)\qquad$ (commutativity, or symmetry). \begin{align*} (x, y) &= \sum_{i=1}^n (x_i + y_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2 \\ &= \sum_{i=1}^n (y_i + x_i)^2 - \sum_{i=1}^n y_i^2 - \sum_{i=1}^n x_i^2 \\ &= (y, x). \end{align*} $\quad\text{(2)}\quad$ $(x, y + z) = (x, y) + (x, z)\qquad$ (distributivity, or linearity).
  \begin{align*} (x, y + z) &= \sum_{i=1}^n (x_i + y_i + z_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n (y_i + z_i)^2 \\ &= \sum_{i=1}^n x_i(y_i + z_i) + \sum_{i=1}^n y_i(x_i + z_i) + \sum_{i=1}^n z_i(x_i + y_i) - 2\sum_{i=1}^n y_iz_i \\ &= 2\sum_{i=1}^n x_iy_i + 2\sum_{i=1}^n x_iz_i \\ &= \left[\sum_{i=1}^n (x_i + y_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2\right] + \left[\sum_{i=1}^n (x_i + z_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n z_i^2\right] \\ &= (x, y) + (x, z) \end{align*} $\quad\text{(3)}\quad$ $c(x, y) = (cx, y) \qquad$ (associativity, or homogeneity). \begin{align*} c(x, y) &= c\sum_{i=1}^n (x_i + y_i)^2 - c\sum_{i=1}^n x_i^2 - c\sum_{i=1}^n y_i^2 \\ &= \sum_{i=1}^n 2cx_iy_i \\ &= \sum_{i=1}^n (cx_i + y_i)^2 - \sum_{i=1}^n (cx_i)^2 - \sum_{i=1}^n y_i^2 \\ &= (cx, y) \end{align*} $\quad\text{(4)}\quad$ $(x, x) > 0 \quad$ if $\quad x \neq O\qquad$ (positivity). \begin{align*} (x, x) &= \sum_{i=1}^n (2x_i)^2 - 2\sum_{i=1}^n x_i^2 \\ &= 2\sum_{i=1}^n x_i^2 \\ &\gt 0 \quad \text{if} \quad x \neq O. \end{align*} As we can see, $(x, y) = \sum_{i=1}^n (x_i + y_i)^2 - \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2$ is an inner product since it satisfies all four axioms. $\quad \blacksquare$