• Linear Algebra: #22 Dual Spaces

Again let V be a vector space over a field F (and, although its not really necessary here, we continue to take F = ℜ or ℂ).

Definition
The dual space to V is the set of all linear mappings f : V → F. We denote the dual space by V^*.

Examples

• Let V = ℜⁿ. Then let f_i be the projection onto the i-th coordinate. That is, if e_j is the j-th canonical basis vector, then
  
  
  
  So each f_i is a member of V^*, for i = 1, . . . , n, and as we will see, these dual vectors form a basis for the dual space.



• More generally, let V be any finite dimensional vector space, with some basis {v₁, . . . , v_n}. Let f_i : V → F be defined as follows. For an arbitrary vector v ∈ V there is a unique linear combination
  
  
  v = a₁v₁ + · · · + a_nv_n . 
  
  Then let f_i(v_i ) = a_i. Again, f_i ∈ V^*, and we will see that the n vectors, f₁, . . . , f_n form a basis of the dual space.



• Let C₀([0, 1]) be the space of continuous functions f : [0, 1] → ℜ. As we have seen, this is a real vector space, and it is not finite dimensional. For each f ∈ C₀([0, 1]) let
  
  
  
  This gives us a linear mapping Λ : C₀([0, 1]) → ℜ. Thus it belongs to the dual space of C₀([0, 1]).



• Another vector in the dual space to C₀([0, 1]) is given as follows. Let x ∈ [0, 1] be some fixed point. Then let Γ_x : C₀([0, 1]) → ℜ is defined to be Γ(f) = f(x), for all f ∈ C₀([0, 1]).



• For this last example, let us assume that V is a vector space with scalar 0 product. (Thus F = ℜ or ℂ) For each v ∈ V, let φ_v(u) = <v, u>. Then φ_v ∈ V^*.

Theorem 56
Let V be a finite dimensional vector space (over ℂ) and let V^* be the dual space. For each v ∈ V, let φ_v : V → ℂ be given by φ_v(u) = <v, u>. Then given an orthonormal basis {v₁, . . . , v_n} of V, we have that {φ_v1, . . ., φ_vn} is a basis of V^*. This is called the dual basis to {v₁, . . . , v_n}.

Proof
Let φ ∈ V^* be an arbitrary linear mapping φ : V → ℂ. But, as always, we remember that φ is uniquely determined by vectors (which in this case are simply complex numbers) φ(v₁), . . . , φ(v_n). Say φ(v_j) ∈ ℂ, for each j. Now take some arbitrary vector v ∈ V. There is the unique expression

Therefore, φ = c₁φ_v1 + · · · + c_nφ_vn, and so {φ_v1, . . ., φ_vn} generates V^*.

To show that {φ_v1, . . ., φ_vn} is linearly independent, let φ = c₁φ_v1 + · · · + c_nφ_vn be some linear combination, where c_j ≠ 0, for at least one j. But then φ(v_j) = c_j ≠ 0, and thus φ ≠ 0 in V^*.

Corollary
dim(V^*) = dim(V).

Corollary
More specifically, we have an isomorphism V → V^*, such that v → φ_v for each v ∈ V.

But somehow, this isomorphism doesn’t seem to be very “natural”. It is defined in terms of some specific basis of V. What if V is not finite dimensional so that we have no basis to work with? For this reason, we do not think of V and V^* as being “really” just the same vector space. [In case we have a scalar product, then there is a “natural” mapping V → V^*, where v → φ_v, such that φ_v(u) = <v, u>, for all u ∈ V.]

On the other hand, let us look at the dual space of the dual space (V^*)^*. (Perhaps this is a slightly mind-boggling concept at first sight!) We imagine that “really” we just have (V^*)^* = V. For let Φ ∈ (V^*)^*. That means, for each φ ∈ V^* we have Φ(φ) being some complex number. On the other hand, we also have φ(v) being some complex number, for each V ∈ V. Can we uniquely identify each V ∈ V with some Φ ∈ (V^*)^*, in the sense that both always give the same complex numbers, for all possible φ ∈ V^*?

Let us say that there exists a v ∈ V such that Φ(φ) = φ(v), for all φ ∈ V^*. In fact, if we define φ_v to be Φ(φ) = φ(v), for each φ ∈ V^*, then we certainly have a linear mapping, V^* → ℂ. On the other hand, given some arbitrary Φ ∈ (V^*)^*, do we have a unique v ∈ V such that Φ(φ) = φ(v), for all φ ∈ V^*? At least in the case where V is finite dimensional, we can affirm that it is true by looking at the dual basis.


Dual mappings 
Let V and W be two vector spaces (where we again assume that the field is ℂ). Assume that we have a linear mapping f : V → W. Then we can define a linear mapping f ^* : W^* → V^* in a natural way as follows. For each φ ∈ W^*, let f ^*(φ) = φ ◦ f. So it is obvious that f ^*(φ) : V → ℂ is a linear mapping. Now assume that V and W have scalar products, giving us the mappings s : V → V^* and t : W → W^*. So we can draw a little “diagram” to describe the situation.

The mappings s and t are isomorphisms, so we can go around the diagram, using the mapping f ^adj = s⁻¹ ◦ f^* ◦ t : W → V. This is the adjoint mapping to f. So we see that in the case V = W, we have that a self-adjoint mapping f : V → V is such that f ^adj = f.

Does this correspond with our earlier definition, namely that <u, f(v)> = <f(u), v> for all u and v ∈ V? To answer this question, look at the diagram, which now has the form

where s(v) ∈ V^* is such that s(v)(u) = <v, u>, for all u ∈ V. Now f ^adj = s⁻¹ ◦ f^* ◦ s; that is, the condition    f ^adj = f becomes s⁻¹ ◦ f^* ◦ s = f. Since s is an isomorphism, we can equally say that the condition is that f^* ◦ s = s ◦ f. So let v be some arbitrary vector in V. We have s ◦ f(v) = f^* ◦ s(v). However, remembering that this is an element of V^*, we see that this means

(s ◦ f(v))(u) = (f^* ◦ s)(v)(u), 

for all u ∈ V. But (s ◦ f(v))(u) = <f(v), u> and (f^* ◦ s)(v)(u) = <v, f(u)>. Therefore we have

<f(v), u> = <v, f(u)>

for all v and u ∈ V, as expected.




This is the last section for this series on Linear Algebra. But that is not to say that there is nothing more that you have to know about the subject. For example, when studying the theory of relativity you will encounter tensors, which are combinations of linear mappings and dual mappings. One speaks of “covariant” and “contravariant” tensors. That is, linear mappings and dual mappings.

But then, proceeding to the general theory of relativity, these tensors are used to describe differential geometry. That is, we no longer have a linear (that is, a vector) space. Instead, we imagine that space is curved, and in order to describe this curvature, we define a thing called the tangent vector space which you can think of as being a kind of linear approximation to the spacial structure near a given point. And so it goes on, leading to more and more complicated mathematical constructions, taking us away from the simple “linear” mathematics which we have seen in this semester.

After a few years of learning the mathematics of contemporary theoretical physics, perhaps you will begin to ask yourselves whether it really makes so much sense after all. Can it be that the physical world is best described by using all of the latest techniques which pure mathematicians happen to have been playing around with in the last few years — in algebraic topology, functional analysis, the theory of complex functions, and so on and so forth? Or, on the other hand, could it be that physics has been loosing touch with reality, making constructions similar to the theory of epicycles of the medieval period, whose conclusions can never be verified using practical experiments in the real world?




IMPORTANT NOTE:
This series on Linear Algebra has been taken from the lecture notes prepared by Geoffrey Hemion. I used his notes when studying Linear Algebra for my physics course and it was really helpful. So, I thought that you could also benefit from his notes. The document can be found at his homepage.

• 9702 November 2012 Paper 22 Worked Solutions | A-Level Physics

Question 1

{Detailed explanations for this question is available as Solution 687 at Physics 9702 Doubts | Help Page 139 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-139.html}

Question 2

{Detailed explanations for this question is available as Solution 990 at Physics 9702 Doubts | Help Page 205 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-205.html}

Question 3

{Detailed explanations for this question is available as Solution 643 at Physics 9702 Doubts | Help Page 128 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-128.html}

Question 4

(a)

Arrangement for producing stationary waves in tube that is closed at one end.

How waves from loudspeaker produce stationary waves in tube:

The waves travel along the tube and are reflected at the closed end (end of tube). The incident and reflected waves are in opposite directions and so, interfere and stationary wave is formed if the tube length is equivalent to λ/4, 3λ/4, etc.

(b)

(i)

Motion of air particles in tube at 1. Point P    2. Point S

1.

No motion of air particle as P is a node (zero amplitude).

2.

Vibration (motion) of the air particles is backwards and forwards / maximum amplitude along length.

(ii)

Speed of sound in tube = 330ms^-1 and frequency of waves from loudspeaker = 880Hz. Calculate length of tube:

L = 3λ/4

v = f λ.                        So, λ = 330 / 880 = 0.375m

L = ¾ x (0.375) = 0.28 (0.281)m

Question 5

{Detailed explanations for this question is available as Solution 685 at Physics 9702 Doubts | Help Page 138 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-138.html}




Question 6

(a)

Hooke’s law states that the extension of a spring is proportional to the force/load applied to it.

(b)

Object M of mass 0.41kg attached to lower end of spring. Spring extends until M is at rest at R.

Spring constant, k = 25Nm^-1. Show that extension is about 0.16m:

F = mg = kx

x = mg / k = 0.41x9.81 / 25 = 0.16m

(c)

Object pulled down a further 0.060m to S and then release. For M, just as it is released:

(i)

Forces acting on M:

Weight and (reaction) force from spring (which is equal to tension in spring)

(ii)

Acceleration of M:

F – weight = ma                                  or (25x0.06) = ma

F = (0.1609 + 0.06) x 25 = 5.52N       or (0.16+0.06) x 25 = 5.5N

a = (5.52 – (0.41x9.81)) / 0.41            or 1.5 / 0.41 and (5.5 – 4.02)

a = 3.7 (3.66)ms^-2                                gives 3.6ms^-2

(d)

Describe energy changes from time object is released to time it first returns to R:

The elastic potential energy / strain energy is converted to kinetic and gravitational potential energy. The stretching / extension reduces while the velocity of object M, along with its height from the ground increases.

Question 7
{Detailed explanations for this question is available as Solution 716 at Physics 9702 Doubts | Help Page 145 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-145.html}

• Linear Algebra: #21 Which Matrices can be Diagonalized?

The complete answer to this question is a bit too complicated. It all has to do with a thing called the “minimal polynomial”.

Now we have seen that not all orthogonal matrices can be diagonalized. (Think about the rotations of ℜ².) On the other hand, we can prove that all unitary, and also all Hermitian matrices can be diagonalized.

Of course, a matrix M is only a representation of a linear mapping f : V → V with respect to a given basis {v₁, . . . , v_n} of the vector space V. So the idea that the matrix can be diagonalized is that it is similar to a diagonal matrix. That is, there exists another matrix S, such that S⁻¹MS is diagonal.

But this means that there must be a basis for V, consisting entirely of eigenvectors.

In this section we will consider complex vector spaces — that is, V is a vector space over the complex numbers ℂ. The vector space V will be assumed to have a scalar product associated with it, and the bases we consider will be orthonormal. We begin with a definition.

Definition
Let W ⊂ V be a subspace of V. Let

 W^⊥ = {v ∈ V : <v, w> = 0, ∀w ∈ W}.

Then W^⊥ is called the perpendicular space to W.

It is a rather trivial matter to verify that W^⊥ is itself a subspace of V, and furthermore W∩ W^⊥ = {0}. In fact, we have:


Theorem 53

V = W ⊕ W^⊥. 

Proof
Let {w₁, . . . , w_m} be some orthonormal basis for the vector space W. This can be extended to a basis {w₁, . . . , w_m, w_m+1, . . . , w_n} of V. Assuming the GramSchmidt process has been used, we may assume that this is an orthonormal basis. The claim is then that {w_m+1, . . . , w_n} is a basis for W^⊥.

Now clearly, since <w_j, w_k> = 0, for j ≠ k, we have {w_m+1, . . . , w_n} ⊂ W^⊥.  If u ∈ W^⊥ is some arbitrary vector in W^⊥, then we have

since <w_j, u> = 0 if j ≤ m. (Remember, u ∈ W^⊥) Therefore, {w_m+1, . . . , w_n} is a linearly independent, orthonormal set which generates W^⊥, so it is a basis. And so we have V = W ⊕ W^⊥.


Theorem 54
Let f : V → V be a unitary mapping (V is a vector space over the complex numbers ℂ). Then there exists an orthonormal basis {v₁, . . . , v_n} for V consisting of eigenvectors under f. That is to say, the matrix of f with respect to this basis is a diagonal matrix.

Proof
If the dimension of V is zero or one, then obviously there is nothing to prove. So let us assume that the dimension n is at least two, and we prove things by induction on the number n. That is, we assume that the theorem is true for spaces of dimension less than n.

Now, according to the fundamental theorem of algebra, the characteristic polynomial of f has a zero, λ say, which is then an eigenvalue for f. So there must be some non-zero vector v_n ∈ V, with f(v_n) = λv_n. By dividing by the norm of v_n if necessary, we may assume that ||v_n|| = 1.

Let W ⊂ V be the 1-dimensional subspace generated by the vector v_n. Then W^⊥ is an n−1 dimensional subspace. We have that W^⊥ is invariant under f. That is, if u ∈ W^⊥ is some arbitrary vector, then f(u) ∈ W^⊥ as well. This follows since

λ<f(u), v_n> = <f(u), λv_n> = <f(u), f(v_n)> = <u, v_n> = 0. 

But we have already seen that for an eigenvalue λ of a unitary mapping, we must have |λ| = 1. Therefore we must have <f(u), v_n> = 0.

So we can consider f, restricted to W^⊥, and using the inductive hypothesis, we obtain an orthonormal basis of eigenvectors {v₁, . . . , v_n-1} for W^⊥. Therefore, adding in the last vector v_n, we have an orthonormal basis of eigenvectors {v₁, . . . , v_n} for V.


Theorem 55
All Hermitian matrices can be diagonalized.

Proof
This is similar to the last one. Again, we use induction on n, the dimension of the vector space V. We have a self-adjoint mapping f : V → V. If n is zero or one, then we are finished. Therefore we assume that n ≥ 2.


Again, we observe that the characteristic polynomial of f must have a zero, hence there exists some eigenvalue λ, and an eigenvector v_n of f, which has norm equal to one, where f(v_n) = λv_n. Again take W to be the one dimensional subspace of V generated by v_n. Let W^⊥ be the perpendicular subspace. It is only necessary to show that, again, W^⊥ is invariant under f. But this is easy. Let u ∈ W^⊥ be given. Then we have

 <f(u), v_n> = <u, f(v_n)> = <u, λv_n> = λ<u, v_n> = λ· 0 = 0

.
The rest of the proof follows as before.

In the particular case where we have only real numbers (which of course are a subset of the complex numbers), then we have a symmetric matrix.

Corollary
All real symmetric matrices can be diagonalized.

Note furthermore, that even in the case of a unitary matrix, the symmetry condition, namely a_jk = a_kj , implies that on the diagonal, we have a_jj = a_jj for all j. That is, the diagonal elements are all real numbers. But these are the eigenvalues. Therefore we have:

Corollary
The eigenvalues of a self-adjoint matrix — that is, a symmetric or a Hermitian matrix — are all real numbers.


Orthogonal matrices revisited
Let A be an n × n orthogonal matrix. That is, it consists of real numbers, and we have A^t = A⁻¹. In general, it cannot be diagonalized. But on the other hand, it can be brought into the following form by means of similarity transformations.

To see this, start by imagining that A represents the orthogonal mapping f : ℜⁿ → ℜⁿ with respect to the canonical basis of ℜⁿ. Now consider the symmetric matrix

B = A + A^t = A + A⁻¹. 

This matrix represents another linear mapping, call it g : ℜⁿ → ℜⁿ, again with respect to the canonical basis of ℜⁿ.

But, as we have just seen, B can be diagonalized. In particular, there exists some vector v ∈ ℜⁿ with g(v) = λg(v), for some λ ∈ ℜ. We now proceed by induction on the number n. There are two cases to consider:

• v is also an eigenvector for f, or 
• it isn’t. 

The first case is easy. Let W ⊂ V be simply W = [v]. i.e. this is just the set of all scalar multiples of v. Let W^⊥ be the perpendicular space to W. (That is, w ∈ W means that <w, v> = 0.) But it is easy to see that W^⊥ is also invariant under f. This follows by observing first of all that f(v) = αv, with α = ±1. (Remember that the eigenvalues of orthogonal mappings have absolute value 1.) Now take w ∈ W^⊥. Then <f(w), v> = α⁻¹<f(w), αv> = α⁻¹<f(w), f(v)> = α⁻¹<fw, v> = α⁻¹ · 0 = 0. Thus, by changing the basis of ℜⁿ to being an orthonormal basis, starting with v (which we can assume has been normalized), we obtain that the original matrix is similar to the matrix

where A^* is an (n−1) ×(n−1) orthogonal matrix, which, according to the inductive hypothesis, can be transformed into the required form.

If v is not an eigenvector of f, then, still, we know it is an eigenvector of g, and furthermore g = f + f ⁻¹. In particular, g(v) = λv = f(v) + f ⁻¹(v). That is,

f(f(v)) = λf(v) − v.

So this time, let W = [v, f(v)]. This is a 2-dimensional subspace of V. Again, consider W^⊥. We have V = W ⊕ W^⊥. So we must show that W^⊥ is invariant under f. Now we have another two cases to consider:

• λ = 0, and 
• λ ≠ 0. 

So if λ = 0 then we have f(f(v)) = −v. Therefore, again taking w ∈ W^⊥, we have <f(w), v> = <f(w), −f(f(v))> = −<w, f(v)> = 0. (Remember that w ∈ W^⊥, so that <w, f(v)> = 0.) Of course we also have <f(w), f(v)> = <w, v> = 0.

On the other hand, if λ ≠ 0 then we have v = λf(v) − f(f(v)) so that <f(w), v> = <f(w), λf(v) − f(f(v))> = λ<f(w), f(v)> − <f(w), f(f(v))>, and we have seen that both of these scalar products are zero. Finally, we again have <f(w), f(v)> = <w, v> = 0.

Therefore we have shown that V = W⊕ W^⊥, where both of these subspaces are invariant under the orthogonal mapping f. By our inductive hypothesis, there is an orthonormal basis for f restricted to the n −2 dimensional subspace W^⊥ such that the matrix has the required form. As far as W is concerned, we are back in the simple situation of an orthogonal mapping ℜ² → ℜ², and the matrix for this has the form of one of our 2 × 2 blocks.

• 9702 June 2013 Paper 22 Worked Solutions | A-Level Physics

Question 1

(a)

SI unit of power:

Power = energy/time = (force x distance/time) = kgm²s^-2 / s

Power = kgm²s^-3

(b)

Turbine that is used to generate electrical power from wind

Power, P = CL²ρv³

Where L is the length of each blade (3 blades) of the turbine, ρ is the density of air, v is the wind speed, C is a constant

(i)

C has no units:

C = P / L²ρv³

Units: L² : m²              ρ = kgm^-3        v³ : m³s^-3

C = kgm²s^-3 / (m² kgm^-3 m³s^-3) = kgm²s^-3 / kgm²s^-3

Thus, the units of the components cancel each other, leading to C having no units.

(ii)

L = 25.0m,      ρ of air = 1.30,            C = 0.931.

Efficiency = 55%        Electric Power output, P = 3.50x10⁵W

Wind speed = v

55% = 3.50x10⁵W

Power available from wind, 100% = 3.50x10⁵ x100/55 = 6.36x10⁵W

P = CL²ρv³

So, v³ = P/ CL²ρ = 6.36x10⁵ / (0.931 x (25)² x 1.3)

v = 9.4ms^-1

(iii)

Reasons why the electrical power output of the turbine is less than the power available from the wind:

Not all kinetic energy of the wind is converted to kinetic energy of the blades / generator. The conversion to electrical energy is also not 100% efficient since heat is produced in the generator / bearing, etc.

Question 2

{Detailed explanations for this question is available as Solution 1089 at Physics 9702 Doubts | Help Page 231 - http://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-231.html}





Question 3

{Detailed explanations for this question is available as Solution 864 at Physics 9702 Doubts | Help Page 173 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-173.html}

Question 4

(a)

Apparatus that demonstrates Brownian motion. Include diagram:

Apparatus: cell with particles e.g. smoke (container must be closed)

Diagram showing suitable arrangements with light illumination and microscope

(b)

Observations made using the apparatus:

Specks/flashes of light are seen to go in random motion.

(c)

2 conclusions about properties of molecules of a gas that follow from the observations:

choose any 2:

We cannot see what is causing the smoke to move hence, the molecules are smaller than the smoke particles.

The continuous motion of the smoke particles implies that motion of the molecules is continuous.

The random motion of the particles implies the random motion of molecules.

Question 5

(a)

Frequency, f =50Hz                transverse wave           speed, v=40ms^-1

(i)

Wavelength of transverse wave on string:

v = fλ

λ = 40/50 = 0.80m

(ii)

How this arrangement may produce a stationary wave on the string:

The waves travel along the string and are reflected at Q / wall / fixed end. The incident and reflected waves interfere / superpose.

(b)

(i)

Label all nodes N and all antinodes A:

Nodes labeled at P, Q and the 2 points at zero displacement

Antinodes labeled at the 3 points of maximum displacement

(ii)

Length of string PQ:

PQ represents 1.5λ. Hence PQ = 0.8 x 1.5 = 1.2m

(iii)

Draw stationary wave at time (t+5.0ms) + explanation:

Period, T = 1/f = 1/50 = 20ms.

5ms is ¼ of cycle

Horizontal line through PQ drawn on Fig 5.2
{The reasoning as to why a horizontal line is obtained has been explained as Solution 41 (c)(iv) at Physics 9702 Doubts | Help Page 7 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-7.html to a question similar to this part. Read it there}

Question 6

(a)

Charge is defined as the product of current with time.

(b)

Resistance, R = 18.0Ω            Power supply of 240V            switched on for 2.60Ms

(i)

Power transformed in heater, P:

P =V²/R = (240)² / 18 = 3200W

(ii)

Current in heater, I:

I = V/R = 240 /18 = 13.3A

(iii)

Charge passing through heater in this time, Q:

Charge, Q = It = 13.3 x 2.6x10⁶ = 3.47 x 10⁷C

(iv)

Number of electrons per second passing a given point in heater:

Number of electrons = 3.47x10⁷ / 1.6x10^-19 = 2.17x10²⁶

Number of electrons per second = 2.17x10²⁶ / 2.6x10⁶ = 8.35x10¹⁹ s^-1

  

Question 7

(a)

(i)

W = 206          X = 82

Y = 4               Z = 2

(ii)

Why mass seems not to be conserved in the reaction:

Mass – energy is conserved. Mass on the right hand side is less because energy is released.

(iii)

Meaning of spontaneous:

The reaction is spontaneous means that the reaction is not affected by external conditions / factors / environment OR 2 examples temperature and pressure.