Inner Product Spaces Examples 1

Recall from the Inner Product Spaces page that if $V$ is a vector space over the field $\mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$), then an inner product defined on $V$ is a function that takes each pair of vectors $u, v \in V$ and maps them to an element in $\mathbb{F}$ that satisfies the following properties:

• $<u, u> ≥ 0$ for all $u \in V$ (Positivity).

• $<u, u> = 0$ if and only if $u = 0$ (Definiteness).

• $<u+v,w> = <u, w> + <v, w>$ for all $u, v, w \in V$ (Additivity in the First Slot).

• $<au, v> = a<u, v>$ for all $a \in \mathbb{F}$ and for all $u, v \in V$ (Homogeneity in the First Slot).

• $<u, v> = \overline{<v, u>}$ for all $u, v \in V$ (Conjugate Symmetry).

We will now look at some more examples of inner products and functions that are not inner products.

Example 1

Consider the vector space $\mathbb{R}^3$. Let $x, y \in \mathbb{R}^3$ where $x = (x_1, x_2, x_3)$ and $y = (y_1, y_2, y_3)$. Define $f(x, y) = x_1y_1 + x_3y_3$. Prove that $f$ is not an inner product for $\mathbb{R}^3$.

Consider the vector $x = (0, b, 0)$ where $b \neq 0$. Then we have that:

(1)

\begin{align} \quad f(x, x) = f((0, b, 0), (0, b, 0)) = 0 \end{align}

We note that $f(x, x) = 0$ but $x = (0, b, 0) \neq (0, 0, 0) = 0$. Therefore $f$ does not satisfy the definiteness property and is hence not an inner product on $\mathbb{R}^3$

Example 2

Prove that if $V$ is an inner product space over the real numbers then for every $u, v \in V$ we have that $<u + v, u - v> = \| u \|^2 - \| v \|^2$.

From the additivity in the first slot we have that:

(2)

\begin{align} \quad <u + v, u - v> = <u, u - v> + <v, u - v> \end{align}

Furthermore, from the additivity in the second slot (which was proved earlier) we have that:

(3)

\begin{align} \quad = <u, u> + <u, -v> + <v, u> + <v, -v> \end{align}

We then apply the property of conjugate homogeneity in the second slot to get that:

(4)

\begin{align} \quad = <u, u> - <u, v> + <v, u> - <v, v> \end{align}

We note that since $V$ is an inner product space over the real numbers then $<v, u> = \overline{<u, v>} = <u, v>$ and so:

(5)

\begin{align} \quad = <u, u> - <u, v> + <u, v> - <v, v> \\ \quad = <u, u> - <v, v> \\ \quad = \| u \|^2 - \| v \|^2 \end{align}