Inner Product Spaces : 2 Marks Important Questions With Answer

INNER PRODUCT SPACEST

2 Marks Important Questions with Answer

1. Define an inner product in vector spaces.

Let V be a vector space over F. An inner product on V is a function that assigns to every ordered pair of vectors x & y in V, a scalar in F, denoted (x, y) such that for all x, y, z in V, & all c in F, the following conditions hold.

1. <x+z,y> = <x, y> + <z, y>

2. <cx, y> = c<x, y>

3.  = <y, x>, where the bar denotes complex conjugate.

4. <x, x> > 0 if x≠0.

Note

(i) For Real numbers (ie F=R), <x,y> = <y, x>

(ii) < ⁿΣ_i=1 a_iv_i,y > = ⁿΣ_i=1 a_i<v_iy> where a_i, ∈ F.

2. Given that U= (2,−3, 6) and V= (8, 2,−3). Find UV?

Solution:

 U.V (16−6−18) = −8.

  U.V exists.

Suppose that U= (1, −3, 0, 5) & V= (3, 6, 4).

Here U.V doesn't exist. Because U is an 1×4 and V is an 1×3 structures.

3. Given that A = . Find the conjugate transpose.

Solution:

4. Show that in F³, the vectors (1, 1, 0), (1, − 1, 1), (− 1, 1, 2) are orthogonal. Are they orthonormal?

Solution:

Let x=(1,1,0); y=(1,−1, 1); z=(−1, 1, 2)

<x,y> = 1−1+0=0;

<x, z> = −1+1+0;

<y, z> = −1−1+2=0

 x, y, z are mutually orthogonal vectors

|| x || = √<x, x> = √[(1)² + (1)² + (0)²] = √2 ≠ 1

|| y || = √<x, y> = √[(1)² + (1)² + (1)²] = √3 ≠ 1.

|| z || = √[(− 1)² + (1)² + (2)² = √6 ≠ 1

The vectors are not orthonormal. If the normalize the vectors in the set, we obtain the orthonormal set as

{( 1/√2, 1/√2, 0 ), ( 1/√3, −1/√3, 1/√3 ), ( −1/√6, −1/√6, 2/√6 ) }

5. Let (a, b) = (c, d) = (0, 1) ∈ R². Prove that < (a, b), (c, d) > = ac – bd is not an inner product.

Solution:

It is given that (a, b) = (c, d) = (0, 1) ∈ R²

Also given that < (a, b), (c, d) > = ac − bd.

 < (0, 1), (0, 1) > = (0)(0) −  (1)(1) = −1.

Hence < (0, 1), (0, 1) > = − 1 < 0.

This is a contradiction to the hypothesis <x, x> > 0 if x ≠ 0.

  It is not an inner product on R².

6. Let β be a basis for a finite dimensional inner product space. If <x, z> = <y,z> for all z ∈ β then prove that x=y.

Solution:

Given that β is a basis and also it is given that <x, y> = <y, z>.

Now we have to prove that x=y

For all z ∈ β, <x, z> = <y, z>

 <x, z> ‒ <y, z> = 0

  <x−y, z> = 0

Here x−y=0 and z = 0. Since z ∈ B, the vector z ≠ 0

 x−y = 0 ⇒ z ≠0.

Hence x−y = 0 ⇒   x=y.

7. Let V=R² and S = {(1, 0), (0, 1)}. Check whether S is orthonormal basis or not.

Solution:

Let V=R².

Define || (a, b) ||= Max {|a|, |b|}

(i) || (a, b) || = Max {|a|,|b|} ≥ 0

(ie) || (a, b) || = 0 if a=0 and b=0

(ii) || c (a, b) || = Max {|ca|, | cb|}

= Max { |c||a|, |c||b|}

= |c| . Max {|a||b|}

= | c | ||(a, b)||

(iii) || (a, b), (c, d) || = || (a+c), (b+d) ||

= Max { |a+c|, |b+d| }

≤ Max { |a|+|c|, |b|+|d| }

≤ Max { |a|,|b|} + Max{|c|,|d||}

Since all the three norm conditions are satisfied,

 || (a, b) || = Max {|a|,|b|} is a norm.

8. Let V=Fⁿ and A ∈ M_m×n(F). Prove that <x, Ay> = <A*x,y> for all x, y ∈ V.

Solution:

Given that V=Fⁿ and A ∈ M_m×n (F).

By the definition of the standard inner product, we have

<x, y> = y*x

 <x, Ay> = (Ay)*x

= y*A*x

= <A*x, y>

 <x, Ay> = <A* x, y>

9. Let V be an inner product space. Let x and y be orthogonal vectors in V. Prove that ||x+y||² = ||x|| + ||y||².

Solution:

Since x and y are orthogonal, <x, y> =0

Then <y, x>==0

|| x + x||² = <x+y, x+y>

= <x, x + y> + <y, x+y>

= <x, x> + <x,y> + <y, x> + <y,y>

= <x, x> + <y, y>

                  ( <x, y> = 0 = <y, x> )

|| x + x||² = || x ||² + || y ||²

10. Prove that || x + y ||² + || x − y ||² = 2|| x ||² + 2|| y ||² for all x, y ∈ V.

Solution:

 || x + y ||² + || x − y ||² = < x + y, x+y> + <x−y, x−y>

 = <x,x+y> + <y, x+y> + <x, x−y> − <y, x−y>

 = 2||x||² + 2||y||²

  || x + y ||² + || x − y ||² = 2|| x ||² + 2||y||² for all x, y ∈ V.

11. Let V be an inner product space. Prove that || x || − || y || ≤ || x − y || for all x, y ∈ V.

Solution:

By the concept of triangle inequality,

We have ||x|| − ||y|| = || x−y+y || − || y ||

 ≤ ||x−y|| + ||y|| − ||y||

 ≤ ||x−y ||

  | || x || − || y || | ≤ || x−y || for all x, y ∈ V.

12. Let R² have the weighted Euclidean inner product defined as <u, v> = 2u₁v₁ + 3u₂v₂ and let u = (1, 1), v = (3, 2), w = (0, −1). Compute the value of <u + v, 3w>.

Solution:

Given that

< u, v> = 2u₁v₁ + 3u₂v₂

 u+v = (1, 1)+(3,2) = (4, 3) and 3w = 3(0,−1) = (0,−3)

 <u+v, 3w> = 2(4)(0) + 3(3)(−3) = 0+(−27) = −27

13. Let V=R². Define T (a, b) = (2a + b, a − 3b) and x = (3, 5). Evaluate T* for V=R².

Solution:

 Let α= {(1, 0), (0, 1)} be the standard basis for V=R².

It is given that T(a, b) = (2a + b, a − 3b)

 T(1, 0) = (2 (1) +0, 1−3 (0)) = (2, 1)

 T(0, 1) = (2 (0)+1, 0−3 (1)) = (1,−3)

Therefore T = 

Since V=R² is real [T]_α = [T]*_α

The adjoint operator T* at x=(3, 5) is defined as follows.

[T*(x)]_α = [T*]_α [x]_α = [T]^*_α [x]_α = [T]_α [x]_α

 T^* at x = (3, 5) is T*(x) = (11, ‒12)

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.

15. 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.

16. State the projection theorem in vector spaces.

Let W be a finite dimensional subspace of an inner product space V and let y ∈ V. Then there exists unique vectors u ∈ W and z ∈ W┴ such that y=u+z. Furthermore, if {v₁, v₂, ..., v_k } is an orthonormal basis for W, then

17. Define g: R² → R by g (a₁, a₂) = 2a₁+a₂. Prove that g is a linear transformation.

Solution

Let β = {e₁, e₂} and let y=g(e₁)e₁ + g(e₂)е₂

= 2e₁ + e₂ = (2, 1)

Then g(a₁, a₂) = < (a₁, a₂), (2, 1) > = 2a₁ + a₂

18. Let T be the linear operator on C² defined by T(a₁, a₂)=(2ia₁+3a₂, a₁−a₂)

If β is the standard ordered basis for C², then find the adjoint.

Solution

 Here T*(a₁, a₂) = (−2ia₁ + a₂, 3a₁−a₂)

19. Let A ∈ Mm×n(F). Then prove that the rank (A*A) = rank (A).

Proof:

By the dimension theorem, we need to show that, for x ∈ Fⁿ, we have A*Ax=0 if and only if Ax=0. Cleanly, Ax=0 ⇒ A* Ax=0. `So assume that A*Ax=0. Then

 0 = <A*Ax, x>_ν = <Ax, A** x>μ = <Ax, Ax>μ ⇒ Ax=0.

20. If A is an m×n matrix such that rank (A)= n, then prove that A*A is invertible.

Proof:

Let A be an m×n matrix and y ∈ F^m. Define W= { Ax: x ∈ Fⁿ };

 (i.e) W=R(L_A). Then there exists a unique vector in W that is closest to y. Call this vector Ax₀ where x₀ ∈ Fⁿ. Then || Ax₀ − y || ≤ || Ax−y || for all x ∈ Fⁿ. So x₀ has the property that E = || Ax₀−y || is minimal.