The Gram Schmidt Orthogonalization: Theorems Part 1

THE GRAM SCHMIDT ORTHOGONALIZATION

THEOREMS PART 1

Theorem 1

Let V be an inner product space and S = { v₁, v₂, ... v_k} be an orthogonal subset of V consisting of non−zero vectors. If y ∈ span(S) then y = .

Solution:

Let y ∈ span (S).

Then y = ^kΣ_i=1 a_iv_i, a_i ∈ F; 1≤i≤k,

We have <y, v_j> =  < ^kΣ_i=1 a_iv_i, v_j >

=   ^kΣ_i=1 a_i <v_i, v_j >

Since <v_i, v_j> = 0 if i ≠ j

<y, v_j> = a_j <v_j, v_j> where i = j.

= a_j ||v_j||²

So a_j = <v_j, v_j> / ||v_j||²

Theorem 2

If S is orthonormal & y ∈ span (S) then

Theorem 3

Let V be an inner product space and let S be an orthogonal subset of V consisting of non−zero vectors then S is linearly independent.

Solution:

Let v₁, v₂, ... v_k ∈ S, and ^kΣ_i=1 a_iv_i = 0.

By previous theorem, with y = 0, we have

a_j = <y, v_j> / ||v_j||²

  = <0, v_j> / ||v_j||²

  a_j = 0 for all j.

  S is linearly independent.

Definition

Fourier Coefficients: Let β be an orthonormal subset of an inner product space V and let x ∈ V. We define the Fourier Coefficients of x relative to β to be the scalars <x, y> where y ∈ β.

Theorem 4

Let V be an inner product space and S = {w₁, w₂,… w_n} be a linearly independent subset of V. Define S' = {v₁, v₂,... v_n }.

where v₁ = w₁ and v_k =  for 2≤k≤n.

Then S' is an orthogonal set of non−zero vectors such that span (S') = span (S).

Proof:

The proof is by mathematical induction on n, the number of vectors in S. For k = 1, 2, ... n, let S_k = { w₁, w₂, …., w_k }

If n = 1, then S₁ = S₁',

since S₁ = {w₁), S'₁={v₁} and w₁ ≠ 0 and v₁ ≠ 0.

 span (S₁) = span (S₁')

If n=2, then S₂ = {w₁, w₂}, S₂' = {v₁, v₂).

where v₁ = w₁, v₂ = w₂ − 

To prove:

S₂' is orthogonal

The set S₂' = {v₁, v₂} is orthogonal.

Moreover if v₂ = 0 then

 w₂ is a linear combination of w₁, which is contradiction to S₂ is linearly independent... v₂ ≠ 0.

 S₂' is an orthogonal set of non zero vectors also,

span (S₂') = span (S₂).

Assume that the set S'_k−1={v₁, v₂, ... v_k−1} is an orthogonal set of non−zero vectors & span (S'_k−1) = span (S_k−1).

To prove:

 S_k' = {v₁, v₂, ... v_k−1, v_k} is an orthogonal set of non−zero vectors where 

W_k ∈ span { v₁, v₂, ..., v_k−1 }

  span (S'_k−1) = span (S_k−1).

W_k ∈ span (S_k−1)

This is a contradiction to S_k which is linearly independent.

  v_k ≠ 0

For 1 ≤ i ≤ k−1

Hence S_k is an orthogonal set of non−zero vectors also

 span (S_k') ≤ span (S_k).

We know that S_k is linearly independent set.

 dim [ span (S_k') ] = dim [ span (S_k) ] = k.

 span (S'_K) = span (S_K)