15.8: Types of Bases

Normalized Basis

Definition: Normalized Basis

A basis \(\left\{b_{i}\right\}\) where each \(b_i\) has unit norm

\[\left\|b_{i}\right\|=1, \quad i \in \mathbb{Z} \nonumber \]

Note

The concept of basis applies to all vector spaces (Section 15.2). The concept of normalized basis applies only to normed spaces (Section 15.3).

You can always normalize a basis: just multiply each basis vector by a constant, such as \(\frac{1}{\left\|b_{i}\right\|}\)

Example \(\PageIndex{1}\)

We are given the following basis:

\[\left\{b_{0}, b_{1}\right\}=\left\{\left(\begin{array}{l}
1 \\
1
\end{array}\right),\left(\begin{array}{c}
1 \\
-1
\end{array}\right)\right\} \nonumber \]

Normalized with \(\ell^{2}\) norm:

\[\begin{array}{c}
\tilde{b}_{0}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1 \\
1
\end{array}\right) \\
\tilde{b}_{1}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1 \\
-1
\end{array}\right)
\end{array} \nonumber \]

Normalized with \(\ell^{1}\) norm:

\[\begin{array}{c}
\tilde{b}_{0}=\frac{1}{2}\left(\begin{array}{c}
1 \\
1
\end{array}\right) \\
\tilde{b}_{1}=\frac{1}{2}\left(\begin{array}{c}
1 \\
-1
\end{array}\right)
\end{array} \nonumber \]

Orthogonal Basis

Orthogonal Basis

a basis {bi}b i in which the elements are mutually orthogonal

∀i,i≠j:(⟨bi,bj⟩=0)i i j b i b j 0

Definition: Orthogonal Basis

A basis \(\left\{b_{i}\right\}\) in which the elements are mutually orthogonal

\[\left\langle b_{i}, b_{j}\right\rangle=0, \quad i \neq j \nonumber \]

Note

The concept of orthogonal basis applies only to Hilbert Spaces (Section 15.4).

Example \(\PageIndex{2}\)

Standard basis for \(\mathbb{R}^2\), also referred to as \(\ell^{2}([0,1])\):

\[\begin{array}{l}
b_{0}=\left(\begin{array}{l}
1 \\
0
\end{array}\right) \\
b_{1}=\left(\begin{array}{l}
0 \\
1
\end{array}\right)
\end{array} \nonumber \]

\[\left\langle b_{0}, b_{1}\right\rangle=\sum_{i=0}^{1} b_{0}[i] b_{1}[i]=1 \times 0+0 \times 1=0 \nonumber \]

Example \(\PageIndex{3}\)

Now we have the following basis and relationship:

\[\left\{\left(\begin{array}{l}
1 \\
1
\end{array}\right),\left(\begin{array}{c}
1 \\
-1
\end{array}\right)\right\}=\left\{h_{0}, h_{1}\right\} \nonumber \]

\[\left\langle h_{0}, h_{1}\right\rangle=1 \times 1+1 \times-1=0 \nonumber \]

Orthonormal Basis

Pulling the previous two sections (definitions) together, we arrive at the most important and useful basis type:

Definition: Orthonormal Basis

A basis that is both normalized and orthogonal

\[\left\|b_{i}\right\|=1, \quad i \in \mathbb{Z} \nonumber \]

\[\left\langle b_{i}, b_{j}\right\rangle \quad, \quad i \neq j \nonumber \]

Notation:

We can shorten these two statements into one:

\[\left\langle b_{i}, b_{j}\right\rangle=\delta_{i j} \nonumber \]

where

\[\delta_{i j}=\left\{\begin{array}{l}
1 \text { if } i=j \\
0 \text { if } i \neq j
\end{array}\right. \nonumber \]

Where \(\delta_{i j}\) is referred to as the Kronecker delta function (Section 1.6) and is also often written as \(\delta[i-j]\).

Orthonormal Basis Example #1

\[\left\{b_{0}, b_{2}\right\}=\left\{\left(\begin{array}{l}
1 \\
0
\end{array}\right),\left(\begin{array}{l}
0 \\
1
\end{array}\right)\right\} \nonumber \]

Orthonormal Basis Example #2

\[\left\{b_{0}, b_{2}\right\}=\left\{\left(\begin{array}{l}
1 \\
1
\end{array}\right),\left(\begin{array}{c}
1 \\
-1
\end{array}\right)\right\} \nonumber \]

Orthonormal Basis Example #3

\[\left\{b_{0}, b_{2}\right\}=\left\{\frac{1}{\sqrt{2}}\left(\begin{array}{l}
1 \\
1
\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}
1 \\
-1
\end{array}\right)\right\} \nonumber \]

Beauty of Orthonormal Bases

Orthonormal bases are very easy to deal with! If \(\left\{b_{i}\right\}\) is an orthonormal basis, we can write for any \(x\)

\[x=\sum_{i} \alpha_{i} b_{i} \nonumber \]

It is easy to find the \(\alpha_i\):

\[\begin{align}
\left\langle x, b_{i}\right\rangle &=\left\langle\sum_{k} \alpha_{k} b_{k}, b_{i}\right\rangle \nonumber \\
&=\sum_{k} \alpha_{k}\left\langle\left(b_{k}, b_{i}\right)\right\rangle
\end{align} \nonumber \]

where in the above equation we can use our knowledge of the delta function to reduce this equation:

\[\begin{array}{c}
\left\langle b_{k}, b_{i}\right\rangle=\delta_{i k}=\left\{\begin{array}{l}
1 \text { if } i=k \\
0 \text { if } i \neq k
\end{array}\right. \\
\left\langle x, b_{i}\right\rangle=\alpha_{i}
\end{array} \nonumber \]

Therefore, we can conclude the following important equation for \(x\):

\[x=\sum_{i}\left\langle\left(x, b_{i}\right)\right\rangle b_{i} \nonumber \]

The \(\alpha_i\)'s are easy to compute (no interaction between the \(b_i\)'s)

Example \(\PageIndex{4}\)

Given the following basis:

\[\left\{b_{0}, b_{1}\right\}=\left\{\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1 \\
1
\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}
1 \\
-1
\end{array}\right)\right\} \nonumber \]

represent \(x=\left(\begin{array}{l}
3 \\
2
\end{array}\right)\)

Example \(\PageIndex{5}\): Slightly Modified Fourier Series

We are given the basis

\[\left.\left\{\frac{1}{\sqrt{T}} e^{j \omega_{0} n t}\right\}\right|_{n=-\infty} ^{\infty} \nonumber \]

on \(L^2([0,T])\) where \(T=\frac{2 \pi}{\omega_0}\).

\[f(t)=\sum_{n=-\infty}^{\infty}\left\langle\left(f, e^{j \omega_{0} n t}\right)\right\rangle e^{j \omega_{0} n t} \frac{1}{\sqrt{T}} \nonumber \]

Where we can calculate the above inner product in \(L^2\) as

\[\left\langle f, e^{j \omega_{0} n t}\right\rangle=\frac{1}{\sqrt{T}} \int_{0}^{T} f(t) \overline{e^{j \omega_{0} n t}} \mathrm{d} t=\frac{1}{\sqrt{T}} \int_{0}^{T} f(t) e^{-\left(j \omega_{0} n t\right)} \mathrm{d} t \nonumber \]

Orthonormal Basis Expansions in a Hilbert Space

Let \(\left\{b_{i}\right\}\) be an orthonormal basis for a Hilbert space \(H\). Then, for any \(x \in H\) we can write

\[x=\sum_{i} \alpha_{i} b_{i} \nonumber \]

where \(\alpha_{i}=\left\langle x, b_{i}\right\rangle\).

• "Analysis": decomposing \(x\) in term of the \(b_i\)

  \[\alpha_{i}=\left\langle x, b_{i}\right\rangle \nonumber \]

• "Synthesis": building \(x\) up out of a weighted combination of the \(b_i\)

  \[x=\sum_{i} \alpha_{i} b_{i} \nonumber \]