Inner Products and Norms: Example Solved Problems - Part 2

INNER PRODUCTS AND NORMS

Example Solved Problems

 

Example 5

For any non−zero vector, x ∈ V prove that y = x / || x || is a vector such that || y || = 1.

Solution:

Consider

||y||² = 1

||y||= 1

 

Note: The process of multiplying a non−zero vector by the reciprocal of its length (norm) is called normalising.

 

Example 6

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

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 ) }

 

Example 7

Let V=C³ with inner product <x,y> =x₁₁+x₂₂+x³₃

where x = (x₁,x₂, x₃) & y = (y₁, y₂, y₃).

Let x=(2,1+i, i), y = (2−i, 2, 1+2i)

Compute (a) (x,y) (b) || x || & || y || (c) || x+y || (d) Verify Cauchy's inequality & Triangle inequality.

Solution:

It is given that <x, y> = x₁₁+x₂₂+x₃₃

(a) <x, y> = 2(2 − ) + 2 (1 + i) + i(1 + 2)

= 2(2+i)+2(1+i)+i(1−2i)

= 4+2i+2+2i+i−2i²

= 4+2i+2i+2+i+2

= 8+5i

(b) ||y|| = √<y,y> =√[(2−i, 2, 1+2i) (2−i, 2, 1+2i)]

= √[(2−i)  +2 (2) + (1 + 2i) ]

= √[(2−i) (2+i) + 2(2) + (1+2i) (1‒2i)]

= √[4+1+4+1+4] = √14

|| x || = √<xx>

√[(2, 1 + i, i) (2, 1 + i, i)] = √[(2)(2) + (1 + i)(1 + ) + i ()]

= √[4 + (1 + i)(1 − i) – î²]

= √[ 4+ (1² − i²) − i² ] = √4+1+1+1 = √7

 (c) ||x+y||² = <x+y, x+y>

= < (4−i, 3+i, 1+3i), (4−i, 3+ i, 1+3i) >

 = (4−i) (4+i) + (3+ i)(3−i) + (1+3i) (1−3i)

= (16−i²) + (9−i²) + (1−9i²) = 17+10+10 = 37

 (d) (i) Cauchy's inequality"

Linear Algebra

| <x,y> | = |8+5i| = √[(8)²+(5)²] = √[64+25] = √89

||x||.||y||= √7 . √14 = √98

 √89 ≤ √98 verified.

(ii) Triangle inequality

|| x + y || ≤ || x || + || y ||

 √37 ≤ √7 + √14

6.08 ≤ 6.39 verified

 

Example 8

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.

 

Example 9

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.

 

Example 10

Let V=R². Define || (a, b) || = Max { | a |, |b| } for (a, b) ∈ V. Prove that it is a norm on vector space V.

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.

 

Example 11

Let A and B be n×n matrices and let c be a scalar.

Then prove that (A + cB)* = A*+

Solution:

Given that A and B are n×n square matrices and c be any scalar.

Now to prove (A + cB)* = A* + B*, let us consider the conjugate transpose of the square matrix A is A*.

 (A*)_ij=

 

Example 12

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

Solution:

Let x=(1, 0) and y = (0, 1).

 <x, y> = (1)(0)+(0)(1)=0+0=0.

|x||=√<x,x> = √1 = 1

|| y || = √<y,y> = √1 = 1

Here ||x|=|| y ||= 1.

 S is an orthonormal basis

 

Example 13

Let S = {,} Verify S is orthonormal basis or not.

Solution:

Let x=  and y= 

To prove S is basis

 = –1/5 – 4/5 = −1 ≠ 0

 S is linearly independent.

Dimension of R² = V is 2.

So S is maximally linearly independent set.

 S is a basis.

S is an orthonormal basis

 

Example 14

Let V be the vector space of polynomial with inner product given by <f,g> = ₀ʃ¹ f(t)g(t) dt. Let f(t) = t + 2, g(t) = t² – 2t − 3.

Find <f,g> & ||ƒ||.

Solution:

Given that <f,g> = ₀ʃ¹ f(t) g(t) dt

 f(t) g(t) = (t + 2) (t² − 2t − 3) = t³−7t−6

< f, g> =  ₀ʃ¹ (t³ − 2t − 3) dt = [ t⁴/4 − 7t²/2 – 6t]¹₀ = [ ¼ − 7/2 − 6]

< f, g> = − 37 / 4.

We know that ||f|| = √<ƒ,ƒ>

 < f, f> = ₀ʃ¹ (t + 2)² dt = ₀ʃ¹ (t² + 4t + 4) dt

= [ t³/3 + 4t²/2 + 4t]¹₀

= [ 1/3 + 4/2 + 4] = 19/3

 ||f|| = √(19/3)

 

Example 15

Consider the set of all continuous complex valued function in [0, 1] and denoted it as V. Let f (t) & g (t) ∈ V. Define <f(t), g(t)> = ₀ʃ¹  f(t)  dt. Verify the conditions of inner product space.

Solution:

 

Example 16

Let V=C³ where C is the set of complex numbers. Defined by <x,y> a₁₁+a₂₂+a₃₃ where x = (a₁, a₃, a₃) & y = (b₁, b₂, b₃). Verify the inner product space.

Solution:

Let x, y, z ∈ V where z= (c₁, c₂, c₃)

 ⇒│a₁│= 0 = |a₂| = |a₃|

  ⇒  x=0.

Conversely x=0, (a₁, a₂, a₃) = 0

⇒  a₁=a₂=a₃ = 0

<x, x> = |a₁|² + |a₂|² + |a₃|² = 0

 V is an inner product space.

 

Example 17

Let V=M_n×n(F) and define <A, B> = trace (B*A) for A, B ∈ V. Verify whether V is an inner product space or not.

 

Solution:

Let A, B, C, ∈ V

(i) <A+B, C> = trace (C* (A+B))

= tr (C*A + C*B)

= tr (C*A) + tr (C*B)

= <A, C> + <B, C>

(ii) <αA, B> = trace (B*(αA))

= α tr (B*A)

= α<A, B>

 (iii)  = [tr (B*A)]* .  =A*_ij

= tr (B*A)*

= tr (BA*)

= tr (A*B)

= <B,A>

(iv) <A, A> = tr (A*A)

If A ≠ 0 then A_ki ≠ 0 for some k_i.

  <A,A> > 0

 

Example 18

Prove that <f(x), g(x)> = ₀ʃ¹ f '(t) g(t) dt on P (R), is not an inner product on the given vector space.

Solution:

Given that < f(x), g(x) > = ₀ʃ¹ f '(t)g(x) dt

Here f(x) and g(x) are in P (R).

Let us consider f(x) = 1

Given that < f(x), g(x) > = ₀ʃ¹ f '(t) g(t) dt

< f(x), f(x) > = ₀ʃ¹  f '(t)f(t) dt

=  ₀ʃ¹ (0)(1) dt = 0                     (f '(x)=0)

This result is contradiction with the result <x, x> > 0 if x ≠ 0.

Hence the given statement is not an inner product.

 

Example 19

Let <f,g> =1/2π  ₀ʃ^2π f(t) dt. Prove that the vector space in continuous complex valued function defined on the interval [0, 2π] is an inner product.

Solution:

 

Example 20

Let V = C{([0,1]). Define <u₁,u₂> =  ₀ʃ^1/2 u₁(t)u₂ dt. Verify whether the given relation is an inner product or not ?

Solution:

Let V = C { [0, 1] }

Now let <u₂,u₁> = ₀ʃ^1/2 u₂(t) • u₁(t) dt

= ₀ʃ^1/2 (i.1) dt = ₀ʃ^1/2 i dt

= i[t]^1/2₀

= i.[1/2]

 <u₂, u₁> = i/2

From equations (1) & (2) we have

 <> ≠  <u₂, u₁>, since ‒i/2 ≠ i/2

 <u₁, u₂>  = ₀ʃ^1/2 u₁(t)u₂(t) dt is not an inner product.

 

Example 21

Let f(t) = t and g(t) = e^t. Find the values of ||f||, |g|| and ||f+g || in C{ [0, 1]}.

Solution:

 Given that f(t) = t and g(t) = et.

 

Example 22

Let V be a vector space over C. Suppose that <•,•> is a real inner product on V, such that [x, ix] = 0. Let <•,•> be the complex valued function defined by <x,y> = [x, y]+i[x, iy] for x,y ∈ V.

Solution:

Let <•,•> be the complex valued function defined by

 <x, y> = [x, y]+i[x, iy] for x, y ∈ V.

(i) <x+z,y> = [x + z, y] + i[x + z, iy]

= [x, y] + [z, y] + i[x, iy] + i[z, iy]

= [x, y] + i[x, iy] + [z, y] + i[z, iy]

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

(ii) For any complex number a+ib ∈ C.

< (a+ib) x, y > = [(a+ib)x, y] + i [(a+ib)x, iy]

= [ax, y] + [ibx, y] + i[ax, iy] + i[ibx, iy]

= a { [x, y]+i [x, iy]) } + ib { [x, y]+i [x, iy]}

= a<x,y> + ib<x,y>

= (a+ib) <x, y>

(iii) 

= [x, y] − i[x, iy]

= [y, x] + i[y, ix]

= <y, x>

 (iv) For x ≠ 0

 <x, x> = [x, x] + i[x, ix]

= [x, x] + 0      

(since [x, ix] = 0)

≥ 0

  <x, y> = [x, y] + i[x, iy] is an inner product

 

Example 23

Let T be a linear operator on an inner product space V and let || T(x) || = || x || for all x. Prove that T is one−to−one.

Solution:

Let V be the vector space and T be a linear operator on V.

Given that || T(x) || = || x || for all x.

To prove T is one to one

Assume that T(x₁) = T (x₂)

Then || T(x₁−x₂) || = ||T(x₁)− T(x₂)|| = ||0|| = 0

                ( T(x₁)=T(x₂) )

  ||T(x₁−x₂) ||= 0

Since ||T(x)|| = ||x|| for all x ∈ V

 || T(x₁ = x₂) || = || x₁−x₂ ||

⇒  || x₁−x₂ || = 0.

And also || x || = 0 if and only if x = 0.

  x₁−x₂=0 ⇒ x₁ = x₂.

  The given linear operator T is one−to−one.

 

Example 24

Let V=C {[0,1]}. Define ||ƒ|| = ₀ʃ¹ |f(t)| dt for all f ∈ V. Prove that it is a norm on the given vector space V.

Solution:

Given that V = C { [0, 1] }

 Defined function is ||f|| = ₀ʃ¹ |f(t)| dt

 (i) ||f|| = ₀ʃ¹ |f(1)|dt ≥ 0 and ||f|| is equal to zero if and only if all the values of f are in [0, 1] is zero.

(ii) ||af|| = ₀ʃ¹ |af(t)| dt = ₀ʃ¹ |a| |f(t)| dt

= |a| ₀ʃ¹ |f(t)| dt

= |a| || ƒ ||

(iii) ||f+ g || = ₀ʃ¹ | (f+g)(t) | dt = ₀ʃ¹ | f(t)+g(t) | dt

≤  ₀ʃ¹ [ |f(t)| + |g(t)| ] dt

= ₀ʃ¹ |f(t)| dt + ₀ʃ¹ |g(t)| dt

= || f || + ||g||

Since all the axioms of the norm are satisfied,

 ||f|| = ₀ʃ¹ | f(t) | dt is a norm on the given vector space V.

 

Example 25

Let . Use Frobenius inner product to find ||A||, || B|| and <A, B>.

Solution:

 

Example 26

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>

 

Example 27

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

 

Example 28

Prove that || x+y ||² + || x − y ||² = 2|| x ||² + 2|| y ||² for all x,y ∈ V (Parallelogram law).

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.

 

Example 29

Let V be the inner product space. Prove that

||x + y||² = ||x||² ± 2Re<x, y> + ||y||² for all x,y ∈ V. Here Re (x, y) denotes the real part of the complex number (x,y).

Solution:

Given that V is the inner product space,

Let us consider ||x+y||² = <x+y, x + y>

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

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

= ||x²|| + <x, y> +  + || y ||²

= ||x||² + 2Re < x,y> + ||y||²

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

Similarly we can prove that

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

 

Example 30

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.