The Gram Schmidt Orthogonalization: Theorems Part 4 - Example Solved Problems

THE GRAM SCHMIDT ORTHOGONALIZATION

Theorems Part 4

WORKED EXAMPLE PROBLEMS

Example 27

Let V=P₃(R) with inner product <f(x), g(x)> = _‒1ʃ¹ f(t)g(t) dt for all f(x), g(x) ∈ V. Compute the orthogonal projection f₁(x) of f(x)=x³ on P₂(R).

Solution:

From Example: 7, we know that

 {v₁, v₂, v₃} = { 1/√2, √(3/2)x, √(5/8), (3x² – 1)} is an orthonormal basis for P₂(R).

 f₁(x) = < f(x), v₁>v₁ + < f(x), v₂> v₂ + <f(x), v₃> v₃

 = 0 + √6/5. √(3/2)x + 0

= 3/5 . x

Example 28

If V=R³ and S = { e₃} then prove that S┴ equals xy – plane.

Solution:

Given that V = R³.

Then { e₁, e₂, e₃ } is a basis for V=R³.

 {e₁, e₂, e₃} = { (1, 0, 0), (0, 1, 0), (0, 0, 1) } and S= {e₃}

 Suppose S┴ = V.

  < v, e₃> =0; v ∈ V.

  (v=α₁е₁ + α₂e₂ + α₃e₃)

 < (α₁e₁ + α₂e₂ + α₃e₃), e₃ > = 0

α₁ <e₁, e₃> + α₂ <e₂, e₃ > + α₃ <e₃, e₃> = 0

 0+0+α₃=0

 α₃ = 0

 S┴=α₁e₁ + α₂e₂          (since α₃=0)

(i.e.) S┴ equals xy-plane.

Example 29

Let V=R². Define u = (2,6) and U = {(x, y); y = 4x}. Compute the orthogonal projection of the given vector on U of the inner product space V and find the distance from V to U.

Solution:

Let V=R² be an inner product space and U be the subspace of V.

It is given that V=R²; u = (2, 6) and U = { (x, y) ; y = 4x }.

We know that the orthogonal projection of u is <u, v₁> v₁

 v₁ = u₁ = (1, 4) since y=4x.

|| v₁ || = √[1² + 4²] = √[1 + 16] = √17

But v₁ = v₁ / || v₁ || = 1/√17 (1, 4)  is a basis for U.

The required orthogonal projection of u is <u, v> v₁.

= 26/17 (1,4)

The required orthogonal projection of u is 29/17 (1, 4).

The distance of the given vector u to be subspace U is the length of u−v.

 Distance = || u−projection of u || = || (2, 6) – 26/17(1, 4) ||

= | | ( 2 – 26/17) (6 – 104/17) ||

= √[ (8/17)² + (–2/17)² ]

= √[4/17] = 2/17

 The distance is 2/√17

Example 30

Find the orthogonal complement of the subspaces S of R⁴ spanned by the two column vectors y₁ and y₂ of the matrix A.

Solution:

Let us consider 

A vector u ∈ R⁴ will be in the orthogonal complement of S if its dot product with the two columns of A, y₁ and y₂ is zero. If we take the transpose of A, then we see that orthogonal complement of S consists of all the vectors u such that A^Tu = 0

The orthogonal complement of S is the null space of the matrix A^T.

 S┴ = N(A^T)

By solving the homogeneous linear systems, we can find a possible for the orthogonal complement which consists of two vectors