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Normalized Basis

Definition 1: Normalized Basis
a basis $b i b i$ where each $b i b i$ has unit norm
$∀ i ,i∈Z:∥ b i ∥=1 i i b i 1$
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Note:
The concept of basis applies to all vector spaces. The concept of normalized basis applies only to normed spaces.
You can always normalize a basis: just multiply each basis vector by a constant, such as $1∥ b i ∥ 1 b i$

Example 1

We are given the following basis: $b 0 b 1 =( 1 1 )( 1 -1 ) b 0 b 1 1 1 1 -1$ Normalized with $ℓ 2 ℓ 2$ norm: $b ~ 0 =12⁢( 1 1 ) b ~ 0 1 2 1 1$ $b ~ 1 =12⁢( 1 -1 ) b ~ 1 1 2 1 -1$ Normalized with $ℓ 1 ℓ 1$ norm: $b ~ 0 =12⁢( 1 1 ) b ~ 0 1 2 1 1$ $b ~ 1 =12⁢( 1 -1 ) b ~ 1 1 2 1 -1$

Orthogonal Basis

Definition 2: Orthogonal Basis
a basis $b i b i$ in which the elements are mutually orthogonal $∀ i ,i≠j:⟨ b i , b j ⟩=0 i i j b i b j 0$
Note:
The concept of orthogonal basis applies only to Hilbert Spaces.

Example 2

Standard basis for $ℝ 2 ℝ 2$, also referred to as $ℓ 2 ⁢ 0 1 ℓ 2 0 1$: $b 0 =( 1 0 ) b 0 1 0$ $b 1 =( 0 1 ) b 1 0 1$ $⟨ b 0 , b 1 ⟩=∑ i =01 b 0 ⁢i⁢ b 1 ⁢i=1×0+0×1=0 b 0 b 1 i 1 0 b 0 i b 1 i 1 0 0 1 0$

Example 3

Now we have the following basis and relationship: $( 1 1 )( 1 -1 )= h 0 h 1 1 1 1 -1 h 0 h 1$ $⟨ h 0 , h 1 ⟩=1×1+1×-1=0 h 0 h 1 1 1 1 -1 0$

Orthonormal Basis

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

Definition 3: Orthonormal Basis
a basis that is both normalized and orthogonal $∀ i ,i∈Z:∥ b i ∥=1 i i b i 1$ $∀ i ,i≠j:⟨ b i , b j ⟩ i i j b i b j$
Notation:
We can shorten these two statements into one: $⟨ b i , b j ⟩= δ i j b i b j δ i j$ where Where $δ i j δ i j$ is referred to as the Kronecker delta function and is also often written as $δ⁢i−j δ i j$.

Example 4: Orthonormal Basis Example #1

$b 0 b 2 =( 1 0 )( 0 1 ) b 0 b 2 1 0 0 1$

Example 5: Orthonormal Basis Example #2

$b 0 b 2 =( 1 1 )( 1 -1 ) b 0 b 2 1 1 1 -1$

Example 6: Orthonormal Basis Example #3

$b 0 b 2 =12⁢( 1 1 )12⁢( 1 -1 ) b 0 b 2 1 2 1 1 1 2 1 -1$

Beauty of Orthonormal Bases

Orthonormal bases are very easy to deal with! If $b i b i$ is an orthonormal basis, we can write for any $xx$

$x=∑i α i ⁢ b i x i α i b i$
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It is easy to find the $α i α i$:
$⟨x, b i ⟩=⟨∑k α k ⁢ b k , b i ⟩=∑k α k ⁢⟨( b k , b i )⟩ x b i k α k b k b i k α k b k b i$
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where in the above equation we can use our knowledge of the delta function to reduce this equation:
$⟨x, b i ⟩= α i x b i α i$
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Therefore, we can conclude the following important equation for $xx$:
$x=∑i⟨(x, b i )⟩⁢ b i x i x b i b i$
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The $α i α i$'s are easy to compute (no interaction between the $b i b i$'s)

Example 7

Given the following basis: $b 0 b 1 =12⁢( 1 1 )12⁢( 1 -1 ) b 0 b 1 1 2 1 1 1 2 1 -1$ represent $x=( 3 2 ) x 3 2$

Example 8: Slightly Modified Fourier Series

We are given the basis $1T⁢ei⁢ ω 0 ⁢n⁢t| n =−∞∞ n 1 T ω 0 n t$ on $L 2 ⁢ 0 T L 2 0 T$ where $T=2⁢π ω 0 T 2 ω 0$. $f⁢t=∑ n =−∞∞⟨(f,ei⁢ ω 0 ⁢n⁢t)⟩⁢ei⁢ ω 0 ⁢n⁢t⁢1T f t n f ω 0 n t ω 0 n t 1 T$ Where we can calculate the above inner product in $L 2 L 2$ as $⟨f,ei⁢ ω 0 ⁢n⁢t⟩=1T⁢∫0Tf⁢t⁢ei⁢ ω 0 ⁢n⁢t¯d t =1T⁢∫0Tf⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t f ω 0 n t 1 T t T 0 f t ω 0 n t 1 T t T 0 f t ω 0 n t$

Orthonormal Basis Expansions in a Hilbert Space

Let $b i b i$ be an orthonormal basis for a Hilbert space $HH$. Then, for any $x∈H x H$ we can write

$x=∑i α i ⁢ b i x i α i b i$
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where $α i =⟨x, b i ⟩ α i x b i$.
• "Analysis": decomposing $x x$ in term of the $b i b i$
$α i =⟨x, b i ⟩ α i x b i$
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• "Synthesis": building $x x$ up out of a weighted combination of the $b i b i$
$x=∑i α i ⁢ b i x i α i b i$
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