Orthonormal Bases of Vector Spaces Examples 1

# Orthonormal Bases of Vector Spaces Examples 1

Recall from the Orthonormal Bases of Vector Spaces page that if $V$ is a finite-dimensional inner product space, then a basis $\{ e_1, e_2, ..., e_n \}$ is said to be an orthonormal basis of $V$ if $<e_i, e_j> = 0$ if $i \neq j$ for $i, j = 1, 2, ..., n$ and $<e_i, e_i> = 1$ for $i = 1, 2, ..., n$. We will now look at some examples problems regarding orthonormal bases of vector spaces.

## Example 1

Prove that the basis $\{ (1, 0), (0, 1) \}$ of the vector space $\mathbb{R}^2$ is orthonormal with the inner product as the dot product.

Let $e_1 = (1, 0)$ and $e_2 = (0, 1)$. Then we have that:

(1)
\begin{align} \quad <e_1, e_1> = <(1, 0), (1, 0)> = 1 \\ \quad <e_1, e_2> = <(1, 0), (0, 1)> = 0 \\ \quad <e_2, e_1> = <(0, 1), (1, 0)> = 0 \\ \quad <e_2, e_2> = <(0, 1), (0, 1)> = 1 \end{align}

Therefore $<e_i, e_j> = 0$ if $i \neq j$ for $i,j = 1, 2$ and $<e_i, e_i> = 1$ for $i = 1, 2$. Thus indeed $\{ (1, 0), (0, 1) \}$ is an orthonormal basis of $\mathbb{R}^2$.

## Example 2

Consider the vector space $\wp_2 (\mathbb{R})$ of polynomials of degree less than or equal to $2$ with real coefficients. Determine whether or not the basis $\{ 1, x, x^2 \}$ of $\wp_2 (\mathbb{R})$ is orthonormal with the inner product defined as $<p(x), q(x)> = \int_0^1 p(x)q(x) \: dx$ for all $p(x), q(x) \in \wp_2 (\mathbb{R})$.

Let $p_1 = 1$, $p_2 = x$, and $p_2 = x^2$. Then we have that:

(2)
\begin{align} \quad <e_1, e_1> = <1, 1> = \int_0^1 1 \: dx = x \biggr \rvert_0^1 = 1 \\ \quad <e_1, e_2> = <1, x> = \int_0^1 x \: dx = \frac{x^2}{2} \biggr \rvert_0^1 = \frac{1}{2} \neq 0 \\ \quad <e_1, e_3> = <1, x^2> = \int_0^1 x^2 \: dx = \frac{x^3}{3} \biggr \rvert_0^1 = \frac{1}{3} \neq 0 \\ \quad <e_2, e_1> = <x, 1> = \int_0^1 x \: dx = \frac{x^2}{2} \biggr \rvert_0^1 = \frac{1}{2} \neq 0 \\ \quad <e_2, e_2> = <x, x> = \int_0^1 x^2 \: dx = \frac{x^3}{3} \biggr \rvert_0^1 = \frac{1}{3} \neq 1 \\ \quad <e_2, e_3> = <x, x^2> = \int_0^1 x^3 \: dx = \frac{x^4}{4} \biggr \rvert_0^1 = \frac{1}{4} \neq 0 \\ \quad <e_3, e_1> = <x^2, 1> = \int_0^1 x^2 \: dx = \frac{x^3}{3} \biggr \rvert_0^1 = \frac{1}{3} \neq 0 \\ \quad <e_3, e_2> = <x^2, x> = \int_0^1 x^3 \: dx = \frac{x^4}{4} \biggr \rvert_0^1 = \frac{1}{4} \neq 0 \\ \quad <e_3, e_3> = <x^2, x^2> = \int_0^1 x^4 \: dx = \frac{x^5}{5} \biggr \rvert_0^1 = \frac{1}{5} \neq 1 \end{align}

Clearly $\{ 1, x, x^2 \}$ is not an orthonormal basis of $\wp_2 (\mathbb{R})$.

## Example 3

Consider the vector space $\mathbb{C}^2$ over the complex numbers ($\mathrm{dim} (\mathbb{C}^2) = 2$). What values of $a \in \mathbb{C}$ (if any) make $\{ (a, 1), (1, -i) \}$ an orthonormal basis of $\mathbb{C}^2$ with respect to the inner product defined as $<x, y> = x_1\bar{y_1} + \bar{x_2}y_2$ for $x, y \in \mathbb{C}^2$?

Let $e_1 = (a, 1)$ and $e_2 = (1, -i)$. For $\{ (a, 1), (1, -i) \}$ to be a basis of $\mathbb{C}^2$ we must find $a$ such that:

(3)
\begin{align} \quad <e_1, e_1> = <(a, 1), (a, 1)> = a\bar{a} + 1\bar{1} = 1 \\ \quad <e_1, e_2> = <(a, 1), (1, -i)> = a \bar{1} + 1 \bar{-i} = 0 \\ \quad <e_2, e_1> = <(1, -i), (a, 1)> = 1 \bar{a} -i \bar{1} = 0 \\ \quad <e_2, e_2> = <(1, -i), (1, -i)> = 1 \bar{1} + -i \bar{i} = 1 \end{align}

Note though that the bottom equation, $<e_2, e_2> = 1\bar{1} + -i\bar{i} = 1 + i^2 = 1 - 1 = 0 \neq 1$, which is independent of what value we choose for $a$, so infact, there is no value of $a$ which make $\{ (a, 1), (1, -i) \}$ an orthonormal basis of $\mathbb{C}^2$ since the vector $(1, -i)$ is not even of unit length!