The Gram Schmidt Orthogonalization: Theorems Part 2 - Example Solved Problems

THE GRAM SCHMIDT ORTHOGONALIZATION

THEOREM PART 2

WORKED EXAMPLE PROBLEMS

Example 13

Let V=P₂(R). Define < f(x), g(x) > = _‒1ʃ¹ f(t)g(t) dt. Prove that the polynomial f (x) = 1 + 2x+3x² as a linear combination of the vectors in the orthonormal basis {u₁, u₂, u₃} for P₂(R).

Solution:

From Example 7

We know that β = { 1/√2, √(3/2)x, √(5/2).(3/2).(x² ‒ 1/3) } is an orthonormal basis for P₂(R).

Let f(x) ∈ P₂(R).

f(x)= ³Σ_i=1 < f(x), ui > ui

 f(x) = < f(x), u₁)u₁ + < f(x), u₂ > u₂ + <f(x), u₃ > u₃

Example 14

Verify the Cauchy - Schwarz inequality for x = (1,−1,3) and y = (2, 0, −1).

Solution:

We know that the Cauchy Schwarz inequality is

 │< x, y> |  ≤ ||x|| ||y||

| x ⋅ y | = | (1, − 1, 3) . (2, 0, − 1) | = | 2+0−3| = |−1| = 1

 || x || = √[(1)² + (−1)² + (3)²] = √[1+1+9] = √11.

 || y || = √[(2)² + (0)² + (− 1)²] = √[4 + 0 + 1] = √5

  | < x, y> | ≤ ||x|| ||y||

 1 ≤ √11.√5

 1 ≤ √55

 Cauchy − Schwarz inequality is satisfied.

Example 15

Show that the following function defines an inner product on R², where u = (u₁, u₂) and v = (v₁, v₂) and

 <u, v> = u₁v₁ + 2u₂v₂.

Solution:

(i) Because the product of real numbers is commutative,

 <u, v> = u₁v₁+2u₂v₂ = v₁u₁+2v₂u₂ = (v, u)

(ii) Let W = (w₁, w₂). Then

 <u, v+w> = u₁ (v₁ + w₁) + 2u₂(v₂+w₂)

= u₁v₁ + u₁w₁ + 2u₂v₂ + 2u₂w₂

= (u₁v₁ + 2u₂v₂) + (u₁w₁ + 2u₂w₂)

= <u, v> + <u, w>

(iii) If c is any scalar, then

 c<u, v> = c(u₁v₁ + 2u₂v₂)

= (cu₁)v_i +2(cu₂)v₂ = <cu, v>

(iv) Because the square of a real number is non negative

 <v, v> = v₁²+2v²₂ ≥ 0

This expression is equal to zero if and only if v=0.

 ( (i.e.) if and only if v₁ = v₂ = 0)

Example 16

Show that the following function is not an inner product on R³, where u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃) defined by

 <u, v> = u₁v₁ − 2u₂v₂ + u₃v₃

Solution:

We can easy prove that

(i) <x,y> = <y, x>

(ii) <x,y+z> = <x,y> + <x, z>

(iii) c<x, y> = <cx, y>

For the condition (iv)

(i.e.) <x, x> ≥ 0 and <x, y> = 0 if and only if x=0

Let x=(1, 2, 1) then

 <x, x> = (1)(1) − 2(2)(2) + (1)(1) = −6 which is less than 0.

  The function is not an inner product.

Example 17

Let f (x) = 1 and g(x)=x be functions in the vector space c [0, 1], with the inner product defined as

 <f,g>= _aʃ^b f(x)g(x) dx

Verify that | <f,g> | ≤ ||ƒ|| || g ||

Solution:

There fore 0.5 ≤ 0.577

(i.e) | <f,g> | ≤ ||f|| ||g||               (0.5≤0.577).

Example 18

Show that the following set is a basis for R⁴.

S={(2, 3, 2,−2), (1, 0, 0, 1), (−1, 0, 2, 1), (−1, 2, −1, 1)}

Solution:

Let v₁ = (2,3,2,−2), v₂ = (1, 0, 0, 1), v₃ = (−1, 0, 2, 1) and v₄= (−1, 2, 1, 1). The set S has 4 non zero vectors. We can show that S is a basis for R⁴ by showing that it is an orthogonal set as follows.

v₁.v₂ = 2+0 +0−2=0

v₁.v₃ =−2+0+4−2=0

v₁.v₄ =−2+6−2−2=0

v₂.v₃ = −1+0+0+1=0

v₂.v₄ =−1+0+0+1=0

v₂.v₄ =1+0−2+1=0

So S is orthogonal and since orthogonal sets are linearly independent it is a basis for R⁴.

Example 19

Apply the Gram Schmidt orthonormalization process to the basis for V=R³ defined by β={(1, 1, 0), (1, 2, 0), (0, 1, 2)}

Solution:

Let y₁ = (1, 1, 0), y₂ = (1, 2, 0) and y₃ = (0, 1, 2)

Let |Δ|=  = 1[4 − 0] − 1[2 − 0] + 0[0 − 0] = 2 ≠ 0.

 {y₁, y₂, y₃} is linearly independent.

Let { x₁, x₂, x₃ } be the orthogonal set.

 {x₁, x₂, x₃} is an orthogonal set.

Let β′ = {x₁,x₂, x₃ } be an orthogonal basis for R³.

Normalizing each vector in β' produces

So β′′ = { u₁, u₂, u₃} is an orthonormal basis in R³.

Example 20

The vectors y₁ = (0, 1, 0) and y₂ = (1, 1, 1) span a plane in R³. By applying Gram−Schmidt orthonormalization process, find an orthonormal basis for this subspace.

Solution:

Let {x₁, x₂} be the orthogonal set.

 {x₁, x₂} be the orthogonal set.

Let β'= {x₁, x₂} be an orthogonal basis for R³.

Normalizing x₁ and x₂ produces the orthonormal set,

 So β" = {u₁, u₂ } is an orthonormal basis for R³.

Example 21

Apply the Gram−Schmidt orthonormalization process to the basis β = {1,x,x²} in P₂, using the inner product

 <p,q> = _-1ʃ¹ p(x) q(x) dx

Solution:

Let β = { 1, x, x² } = { x₁, x₂, x₃ }

 Let x₁ = y₁ = 1

= x² ‒ 1/3

Now by normalizing β' = { x₁, x₂, x₃} we have

Example 22

Let R³ have the inner product space. Use Gram−Schmidt process to obtain the orthogonal basis β for span (S) and normalize the vectors in this basis to obtain an orthonormal basis β where

 S = {(1, 0, 1), (0, 1, 1), (1, 3, 3)} and x=(1, 1, 2).

Solution:

Let us consider S={ u₁, u₂, u₃ }

where u₁ = (1, 0, 1), u₂ = (0, 1, 1) and u₃ = (1, 3, 3)

Let v₁ = u₁ = (1, 0, 1)

Now <u₂, v₁> = < (0, 1, 1), (1, 0, 1) > = 0+0+1 = 1

 ||v₁|| = √[1² + 0² + 1²] = √2

 v₂ = (0, 1, 1)   ‒  1/2(1, 0, 1)

= (−1/2, 1, 1/2)

 v₂ = (−1/2, 1, 1/2)

Now v₃ = 

 <u₃, v₁> = < (1, 3, 3), (1, 0, 1) > = 1+0+3 = 4

  < u₃, v₂ > = < (1, 3, 3), ( −1/2, 1, 1/2) > = −1/2 + 3 + 3/2 = 4

form an orthogonal basis on R³.

The norms of these vectors are respectively.

|| v₁ || = √2, || v₂ || = √(3/2) and || v₃ || = √[ 1/9 + 1/9 + 1/9 ] = √(1/3)

  The orthonormal basis for R³ is

The orthonormal basis for R³ is β ={y₁, y₂, y₃ }

 β= { ( 1/√2, 0, 1/√2), ( ‒1/√6, 2/√6, 1/√6), ( 1/√3, 1/√3, ‒1/√3) }

Example 23

Let V=R⁴. Let S = {u₁, u₂, u₃}. Define u₁ = (1,−2,− 1, 3), u₂ = (3, 6, 3, − 1), u₃ = (1, 4, 2, 8) and x = (− 1, 2, 1, 1). Use Gram−Schmidt process to compute the orthogonal vectors and normalize these vectors in this basis to form an orthonormal basis.

Solution:

Given that S = { u₁, u₂, u₃ }

where u₁ = (1, −2, −1, 3), u₂ = (3, 6, 3,−1) and u₃ = (1, 4, 2, 8).

Let v₁ = u₁ = (1, −2, −1, 3).

Now <u₂, v₁> = < (3, 6, 3, 1), (1, −2, −1, 3) > = 3−12−3−3 = −15.

  || v₁ || = √ [ 1² + (− 2)² + (− 1)² + (3)² ] = √[ 1 +4 +1 + 9 ] = √15

v₂ = (3, 6, 3, − 1) − (−15/15)(1, − 2, − 1, 3)

v₂ = (3, 6, 3, 1) + (1, −2, −1, 3) = (4, 4, 2, 2)

Now v3 = 

 < u₃, v₁ > = < (1, 4, 2, 8), (1, − 2, − 1, 3) > = 1−8−2 + 24 = 15

  < u₃, v₂ > = < (1, 4, 2, 8), (4, 4, 2, 2) > = 4+16+4+16 = 40

 || v₂ || = √ [ 4² + 4² + 2² + 2² ] = √[ 16 + 16 + 4 + 4 ] = √40

  v₃ = (1, 4, 2, 8) − 15/15(1, −2, −1, 3) – 40/40(4, 4, 2, 2)

= (1, 4, 2, 8) + (−1, 2, 1, −3) + (−4,−4,−2,−2)

  v₃ = (− 4, 2, 1, 3)

Thus v₁ = (1,−2, 1, 3), v₂ = (4, 4, 2, 2) and v₃ = (−4, 2, 1, 3) form an orthogonal basis for R³.

The norms of these vectors are

 ||v₁|| = √15, ||v₂|| = √40 and ||v₃|| = √[ (− 4)² + 2² + 1² + 3² ] = √30.

  The orthonormal basis for R⁴ is β = {y₁, y₂, y₃ } where y₁, y₂ and y₂ are

 y₁ = v₁ / ||v₁|| = 1/√15  (1, −2, −1, 3) = (  1/√15, ‒2/√15, ‒1/√15, 3/√15 )

y₂ = v₂ / ||v₂|| = 1/√40  (4, 4, 2, 2) = (  4/√40, 4/√40, 2/√40, 2/√40 )

y₃ = v₃ / ||v₃|| = 1/√30  (‒4, 2, 1, 3) = (  ‒4/√30, 2/√30, 1/√30, 3/√30 )

 Example 24

Let V=span (S). Define <f,g> = ₀ʃ^π ƒ (t) g(t) dt,

S = {sin t, cost, 1, t} and h(t)=2t+1. Use Gram−Schmidt process to compute the orthogonal vectors and normalize these vectors to an orthonormal basis.

Solution:

Let V = span (S) and S={u₁, u₂, u₃, u₄}. It is given that <f,g> = ₀ʃ^π f(t)g(t) dt, S = { sin t, cos t, 1, t } and h(t) = 2t + 1.

Let u₁ = sint, u₂ = cost, u₃= 1 and u₄ =t.

Now let v₁ = u₁ = sin t.

By using integration by parts for the R.H.S

Let u=t, dv=sin t dt, then du = dt, v=−cos t

We know that ʃudv = uv − ʃvdu

Again by applying integration by parts for the above integration

Let u=t, dv = cost dt then du= dt, v sin t.

 v₁ = sint, v₂ = cos t, v₃ = 1/π [π − 4sin t] and v₄ = 1/2π [2πt + 8cost − π²]

These are the orthogonal basis.

The norms of these vectors are

||v₁|| = √sin t;

|| v₂ || = √cost;

||v₃|| = √ ( 1/π [π‒ 4sin t] ) and

|| v₄ || = √ ( 1/2π [ 2πt + 8cos t ‒ π²] )

The orthogonal basis of these vectors are β= {y₁, y₂, y₃, y₄ }.

Where

y₁ = v₁ / ||v₁|| ;

y₂ = v₂ / ||v₂|| ;

y₃ = v₃ / ||v₃|| ;

y₄ = v₄ / ||v₄|| ;

Example 25

Let V=c[0,1]. Define <f,g> = ₀ʃ¹ f(t)g(t) dt. Let S = {u₁, u₂ } where u₁ = t and u₂= √t. Apply Gram−Schmidt process to compute the orthogonal vectors and normalize these vectors to an orthonormal basis.

Solution:

It is given that <f,g> =  ₀ʃ¹ f(t)g(t) dt

Also it is given that S = { u₁, u₂ } = { t, √t }, u₁ =t, and u₂ = √t.

Let v₁ = u₁ =t.

 v₁ = t and v₂ = √t – 6/5t form an orthogonal basis.

The norms of these vectors are obtained as

The corresponding norms are given by

  ||v₁|| = 1/√3 and ||v₂||  = 1 / √50

 The orthonormal basis is β = {y₁, y₂}