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Homework for MATH6106
March 25, 2010
1
Contents
1 Topics in Three Dimensional Geometry 31.1 Assignment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Solutions to Assignment 1 . . . . . . . . . . . . . . . . . . . . 5
2 Partial Derivatives 102.1 Assignment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Solutions to Assignment 2 . . . . . . . . . . . . . . . . . . . . 11
3 Vector Calculus 143.1 Assignment 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2 Solutions to Assignment 3 . . . . . . . . . . . . . . . . . . . . 16
4 Linear Equations, Linear Independence and Gram Schmidt 194.1 Assignment 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Solutions to Assignment 4 . . . . . . . . . . . . . . . . . . . . 20
5 Matrices 285.1 Assignment 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 285.2 Solutions to Assignment 5 . . . . . . . . . . . . . . . . . . . . 29
6 Eigenvalues, Eigenvectors and Diagonalization 306.1 Assignment 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.2 Solutions to Assignment 6 . . . . . . . . . . . . . . . . . . . . 31
7 Operators 337.1 Assignment 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 337.2 Solutions to Assignment 7 . . . . . . . . . . . . . . . . . . . . 34
8 Fourier Series 398.1 Assignment 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 398.2 Solutions to Assignment 8 . . . . . . . . . . . . . . . . . . . . 40
1 Topics in Three Dimensional Geometry
1.1 Assignment 1
(1) Consider three points in three dimensional space; A = (1, 1, 1), B =(1, 0,−1) and C = (0, 1, 2).
(i) Find vectors ~AB, ~AC and ~BC.
(ii) Find ~AB × ~AC.
(iii) Write an equation for the plane which contains the points A, B,C.
(iv) Write an equation for the line passing through point A and parallelto vector BC.
(2) *Let A = (2, 3,−3), let line L1 be given by x = 2−t, y = 2+t, z = −4+3t,and let line L2 be x = 7 + 2t, y = −t, z = 4 + t. Find the equations for:
(i) the line through A parallel to L1
(ii) the line through A perpendicular to both L1 and L2
(iii) the plane through A perpendicular to L1
(iv) the plane through A parallel to L1 and L2
(3) Show that the point (−1, 1,√
3) lies on the sphere x2 + y2 + z2 = 5 andfind an equation of the plane tangent to the sphere at the point.
(4) Express the following equations in rectangular coordinates.(i) r = 3 (ii) r2 = z2
(iii) ρ = 3 (iv) ρ cosφ = 2
(5) Express the following functions in the specified coordinate system.
(i) In polar coordinates:
f(x, y) = (x2 − y2)2 + 4x2y2.
(ii) In cylindrical coordinates:
g(x, y, z) = x2 + y2 + 2y/x+ tan z.
(iii) In spherical coordinates:
h(x, y, z) =z
x2 + y2 + z2+ x2 + y2.
(6) Find ∂z∂x
and ∂z∂y
if
(a) z = x2 + 2x3y2 (c) z = xx+y
(b) z = xe−y (d) z = x(2x+ 5y)4
(7) Find ∂2f∂y2 and ∂2f
∂x∂yif
(a) f(x, y) = ln(2x+ 3y)
(b) f(x, y) = x+yx−y
(8) Find
(a) ∂3(ex cos y)∂y3 (c) ∂3(x sin y)
∂x3
(b) ∂3(ex cos y)∂x3 (d) ∂3(x sin y)
∂x∂y2
(9) If x = ρ sin(φ) cos(θ)
(a) ∂2x∂ρ2
(b) ∂2x∂φ∂θ
(10) Consider the function u(x, y) = xyx+y
. Show that
x2∂2u
∂x2+ 2xy
∂2u
∂x∂y+ y2∂
2u
∂y2= 0.
1.2 Solutions to Assignment 1
1. Consider three points in three dimensional space; A = (1, 1, 1), B =(1, 0,−1) and C = (0, 1, 2).
(i) The vectors are:
~AB = ~OB − ~OA
= (1, 0,−1)− (1, 1, 1)
= (0,−1,−2)
~AC = ~OC − ~OA
= (0, 1, 2)− (1, 1, 1)
= (−1, 01)
~BC = ~OC − ~OB
= (0, 1, 2)− (1, 0,−1)
= (−1, 1, 3)
(ii)
~AB × ~AC =
∣∣∣∣∣∣~i ~j ~k0 −1 −2−1 0 1
∣∣∣∣∣∣= ((−1)(1)− (0)(−2))~i− ((0)(1)− (−2)(−1))~j + ((0)(0)− (−1)(−1))~k
= −~i+ 2~j − ~k
(iii) A plane which contains the points A, B and C, will contain the
vectors ~AB and ~AC. Therefore by properties of the cross product~AB × ~AC = (−1, 2,−1) will be a normal vector to the plane. An
equation of the plane, in point-normal form, choosing A as a pointon the plane and ~n = (−1, 2,−1) as a normal vector, is:
(−1)(x− 1) + (2)(y − 1) + (−1)(z − 1) = 0.
(iv) We can write a vector equation for the line, as we have a point on the
lineA = (1, 1, 1) and a vector parallel to the line ~v = ~BC = (−1, 1, 3):
A+ t~v = ~r
(1, 1, 1) + t(−1, 1, 3) = (x, y, z)
Parametric equations for the line are:
x = 1− t
y = 1 + t
z = 1 + 3t
2. L1 is parallel to vector ~u = (−1, 1, 3); L2 is parallel to vector ~v = (2,−1, 1)
(i) Parametric equations for the line:x = 2− t, y = 3 + t, z = −3 + 3t
(ii) ~u×~v = (4, 7,−1) is parallel to the required line. Parametric equationsfor the ine are: x = 2 + 4t, y = 3 + 7t, z = −3− t.
(iii) ~u is normal to the plane; equation of the plane is:
−(x− 2) + y − 3 + 3(z + 3) = 0
−x+ y + 3z = −8
(iv) ~u× ~v = (4, 7,−1) is normal to plane. Equation of the plane is:
4(x− 2) + 7(y − 3)− (z + 3) = 0
4x+ 7y − z = 32
3.(−1)2 + (1)2 + (
√3)2 = 5.
Therefore (−1, 1,√
3) satisfies the equation of the sphere and hence lieson the sphere.The radius vector ~OP = (−1, 1,
√3) is perpendicular to the tangent plane
at the point P = (−1, 1,√
3). Therefore the tangent plane has normalvector ~n = (−1, 1,
√3) and contains the point (−1, 1,
√3). Hence equation
in point-normal form is:
−(x+ 1) + y − 1 +√
3(z −√
3) = 0
−x+ y +√
3z = 5
4. (i) Cylinder with radius 3; x2 + y2 = 9 or√x2 + y2 = 3
(ii) Cone; x2 + y2 = z2
(iii) Sphere with radius 3; x2 + y2 + z2 = 9 or√x2 + y2 + z2 = 3
(iv) Plane; z = 2
5. Express the following functions in the specified coordinate system.
(i) In polar coordinates:f(x, y) = (x2 − y2)2 + 4x2y2.
f(r, θ) = r4.
(ii) In cylindrical coordinates:g(x, y, z) = x2 + y2 + 2y/x+ tan z.
g(r, θ, z) = r2 + 2 tan θ + tan z.
(iii) In spherical coordinates:h(x, y, z) = zx2+y2+z2 + x2 + y2.
h(ρ, θ, φ) =cosφ
ρ+ ρ2 sin2(φ).
6. Find ∂z∂x
and ∂z∂y
if
(a)
z = x2 + 2x3y2
∂z
∂x= 2x+ 6x2y2
∂z
∂y= 4x3y
(b)
z = xe−y
∂z
∂x= e−y
∂z
∂y= −xe−y
(c)
z =x
x+ y∂z
∂x=
y
(x+ y)2
∂z
∂y=
−x(x+ y)2
(d)
z = x(2x+ 5y)4
∂z
∂x= 8x(2x+ 5y)3 + (2x+ 5y)4
∂z
∂y= 20x(2x+ 5y)3
7. Find ∂2z∂y2 and ∂2z
∂x∂yif
(a) z = ln(2x+ 3y)
∂2z
∂y2=
−9
(2x+ 3y)2
∂2z
∂x∂y=
−6
(2x+ 3y)2
(b) z = x+yx−y
∂2z
∂y2=
4x
(x− y)3
∂z
∂x∂y=
−2(x+ y)
(x− y)3
8. (a) ∂3ex cos(y)∂y3 = ex sin y
(b) ∂3ex cos(y)∂x3 = ex cos y
(c) ∂3ex sin(y)∂x3 = 0
(d) ∂3x sin(y)∂x∂y2 = − sin y
9. If x = ρ sin(φ) cos(θ)
(a) ∂2x∂ρ2 = 0
(b) ∂2x∂φ∂θ
= −ρ cos(φ) sin(θ)
10. Consider the function u(x, y) = xyx+y
. Show that
x2∂2u
∂x2+ 2xy
∂2u
∂x∂y+ y2∂
2u
∂y2= 0.
Proof.
∂u
∂x=
(x+ y)y − xy
(x+ y)2=
y2
(x+ y)2
∂u
∂y=
(x+ y)x− xy
(x+ y)2=
x2
(x+ y)2
∂2u
∂x2=
∂
∂x(
y2
(x+ y)2) = y2 (−2)
(x+ y)3=
−2y2
(x+ y)3
∂2u
∂y2=
∂
∂y(
x2
(x+ y)2) = x2 (−2)
(x+ y)3=
−2x2
(x+ y)3
∂2u
∂x∂y=
∂
∂x(
x2
(x+ y)2) =
(x+ y)2(2x)− x2(2)(x+ y)
(x+ y)4
=(x+ y)[2x(x+ y)− 2x2]
(x+ y)4
=2x2 + 2xy − 2x2
(x+ y)3
=2xy
(x+ y)3
x2∂2u
∂x2+2xy
∂2u
∂x∂y+y2∂
2u
∂y2= x2 (−2y2)
(x+ y)3+(2xy)
(2xy)
(x+ y)3+y2 (−2x2)
(x+ y)3= 0.
2 Partial Derivatives
2.1 Assignment 2
1. Let f = f(x, y), x = r cos θ, y = r sin θ.
(i) Find ∂x∂r, ∂x
∂θ, ∂y
∂r, ∂y
∂θ.
(ii) Use the chain rule to determine ∂f∂θ, ∂
∂r(∂f
∂x), ∂
∂r(∂f
∂y).
(iii) Use (ii) to show that
∂2f
∂r∂θ= −r sin θ cos θ
∂2f
∂x2+r sin θ cos θ
∂2f
∂y2+(r cos2 θ−r sin2 θ)
∂2f
∂x∂y−sin θ
∂f
∂x+cos θ
∂f
∂y.
2. Find and classify the critical points of the following functions.
(i) f(x, y) = 4 + x3 + y3 − 3xy
(ii) f(x, y) = x3 − x+ xy2
2.2 Solutions to Assignment 2
1. Let f = f(x, y), x = r cos θ, y = r sin θ. Show that,
∂2f
∂r∂θ= −r sin θ cos θ
∂2f
∂x2+r sin θ cos θ
∂2f
∂y2+(r cos2 θ−r sin2 θ)
∂2f
∂x∂y−sin θ
∂f
∂x+cos θ
∂f
∂y.
Proof. (i)
f = f(x, y)
x = r cos θ
y = r sin θ∂x
∂r= cos θ
∂x
∂θ= −r sin θ
∂y
∂r= sin θ
∂y
∂θ= r cos θ
(ii)
∂f
∂θ=
∂f
∂x
∂x
∂θ+∂f
∂y
∂y
∂θ= −r sin θ
∂f
∂x+ r cos θ
∂f
∂y(1)
∂
∂r(∂f
∂x) =
∂2f
∂x2
∂x
∂r+
∂2f
∂y∂x
∂y
∂r= cos θ
∂2f
∂x2+ sin θ
∂2f
∂y∂x(2)
∂
∂r(∂f
∂y) =
∂2f
∂x∂y
∂x
∂r+∂2f
∂y2
∂y
∂r= cos θ
∂2f
∂x∂y+ sin θ
∂2f
∂y2(3)
(iii)
∂2f
∂r∂θ=
∂
∂r(∂f
∂θ)
=∂
∂r(−r sin θ
∂f
∂x+ r cos θ
∂f
∂y)
= − sin θ[∂
∂r(r∂f
∂x)] + cos θ[
∂
∂r(r∂f
∂y)], Since sin θ and
cos θ are constant when we differentiate with respect to r.
= − sin θ[r∂
∂r(∂f
∂x) +
∂f
∂x] + cos θ[r
∂
∂r(∂f
∂y) +
∂f
∂y], Using the product rule.
= −r sin θ[∂
∂r(∂f
∂x)]− sin θ
∂f
∂x+ r cos θ[
∂
∂r(∂f
∂y)] + cos θ
∂f
∂y
= − sin θ∂f
∂x+ cos θ
∂f
∂y
−r sin θ[cos θ∂2f
∂x2+ sin θ
∂2f
∂y∂x]
+r cos θ[cos θ∂2f
∂x∂y+ sin θ
∂2f
∂y2], Using equations (2) and (3).
= −r sin θ cos θ∂2f
∂x2+ r sin θ cos θ
∂2f
∂y2
+(r cos2 θ − r sin2 θ)∂2f
∂x∂y− sin θ
∂f
∂x+ cos θ
∂f
∂y
2. Find and classify the critical points of the following functions.
(i) f(x, y) = 4 + x3 + y3 − 3xyCriticial Points ={(0,0),(1,1)}
(0, 0) : Saddle Point
(1, 1) : Local minimum
(ii) f(x, y) = x3 − x+ xy2
Criticial Points ={(0, 1), (0,−1), ( 1√3, 0), (− 1√
3, 0)}
(0, 1) : Saddle Point
(0,−1) : Saddle Point
(1/√
3, 0) : Local minimum
(−1/√
3, 0) : Local maximum
3 Vector Calculus
3.1 Assignment 3
1. Consider the function f(xy) = 2x2 + 3xy + xy2 − y3.
(i) Find ∇f .
(ii) At the point P (0, 1) find the rate of increase of f in the directionof the vector ~v =~i+~j.
(iii) If f represents the height of a landscape and you are standing onthe ground above P (0, 1) and wish to walk uphill as steeply aspossible, which way should you walk? Give your answer as a unitvector.
2. Consider the function f(x, y) = xexy. The graph of f is given by theequation z = xexy. The point P (2, 0, 2) lies on the graph.
(i) Find a vector field which is perpendicular to the surface z = xexy.
(ii) Find a normal vector to the surface at the point P (2, 0, 2).
(iii) Find an equation for the tangent plane at the point P (2, 0, 2).
Hint: For this question you may like to use the fact that the gradientis perpendicular to level sets.
3. For the vector functions
(a) ~v(x, y, z) = (x2 + y2, 2xy, xyz)
(b) ~F (x, y, z) = yz2~i+ xy~j + yz~k
(i) Calculate the divergence.
(ii) Calculate the curl.
4. Calculate the Laplacian of the following functions, using the appropri-ate formulae (given below) depending on the coordinate system usedto define the function.
(i) f(x, y, z) = 3x2 + y2 + sin z
(ii) f(r, θ) = r cos θ
(iii) f(r, θ, z) = z2 − r2
(iv) f(ρ, θ, φ) = ρ2 sin2 θ sin2 φ
Note. The formulae for the Laplacian:
Rectangular : ∇2f =∂2f
∂x2+∂2f
∂y2+∂2f
∂z2
Polar : ∇2f =1
r
∂
∂r
(r∂f
∂r
)+
1
r2
∂2f
∂θ2
Cylindrical : ∇2f =1
r
∂
∂r
(r∂f
∂r
)+
1
r2
∂2f
∂θ2+∂2f
∂z2
Spherical : ∇2f =1
ρ2
∂
∂ρ
(ρ2∂f
∂ρ
)+
1
ρ2 sin θ
∂
∂θ
(sin θ
∂f
∂θ
)+
1
ρ2 sin2 θ
∂2f
∂φ2
3.2 Solutions to Assignment 3
1. Consider the function f(xy) = 2x2 + 3xy + xy2 − y3.
(i) ∇f = (4x+ 3y + y2, 3x+ 2xy − 3y2).
(ii) At the point P (0, 1) find the rate of increase of f in the directionof the vector ~v =~i+~j.We are asked to find the directional derivative of f at the pointP in the direction ~v.
∇f = (4x+ 3y + y2, 3x+ 2xy − 3y2)
∇f(0, 1) = (4,−3)
D~vf(P ) =∇f(P ).~v
|~v|
=(4,−3).(1, 1)√
12 + 12
= 1/√
2
(iii) If f represents the height of a landscape and you are standing onthe ground above P (0, 1) and wish to walk uphill as steeply aspossible, which way should you walk? Give your answer as a unitvector.We know that
– ∇f points in the steepest direction.
– |∇f | gives the slope in the steepest direction.
Therefore we should walk in the direction 1√42+(−3)2
4~i − 3~j =
45~i− 3
5~j.
2. Consider the function f(x, y) = xexy. The graph of f is given by theequation z = xexy. The point P (2, 0, 2) lies on the graph.
(i) Find a vector field which is perpendicular to the surface z = xxy.We know that the gradient of a function is perpendicularto its level sets. We use this fact to find a vector field that isperpendicular to the surface z = xexy.Define a new function
g(x, y, z) = xexy − z.
Then our surface is the level set
g(x, y, z) = 0
xexy − z = 0
z = xexy
The gradient of g is perpendicular to the level set g(x, y, z) = 0.
∇g = (∂g
∂x,∂g
∂y,∂g
∂z)
= (xyexy + exy, x2exy,−1)
(ii) Find a normal vector to the surface at the point P (2, 0, 2).Since the gradient function gives a vector field which is perpen-dicular to the surface, ∇g(2, 0, 2) is a normal vector to the surfaceat the point P (2, 0, 2).
∇g = (∂g
∂x,∂g
∂y,∂g
∂z)
= (xyexy + exy, x2exy,−1)
∇g(2, 0, 2) = (1, 4,−1)
(iii) Find an equation for the tangent plane at the point P (2, 0, 2).To write the equation of a plane we need a normal vector and apoint on the plane. (1, 4,−1) is a normal vector to the plane andP (2, 0, 2) is a point on the plane. Therefore the equation of thetangent plane:
1(x− 2) + 4(y − 0)− 1(z − 2) = 0.
3. For the vector functions
(a) ~v(x, y, z) = (x2 + y2, 2xy, xyz)div ~v = 4x+ xycurl ~v = xz~i− yz~j
(b) ~F (x, y, z) = yz2~i+ xy~j + yz~kdiv ~v = x+ ycurl ~v = z~i+ 2yz~j + (y − z2)~k
4. Calculate the Laplacian of the following functions, using the appropri-ate formulae (given below) depending on the coordinate system usedto define the function.
(i) f(x, y, z) = 3x2 + y2 + sin z
∇2f =∂2f
∂x2+∂2f
∂y2+∂2f
∂z2= 8− sin z
(ii) f(r, θ) = r cos θ
∇2f =1
r
∂
∂r
(r∂f
∂r
)+
1
r2
∂2f
∂θ2= 0
(iii) f(r, θ, z) = z2 − r2
∇2f =1
r
∂
∂r
(r∂f
∂r
)+
1
r2
∂2f
∂θ2+∂2f
∂z2= 6
(iv) f(ρ, θ, φ) = ρ2 sin2 θ sin2 φ
∇2f =1
ρ2
∂
∂ρ
(ρ2∂f
∂ρ
)+
1
ρ2 sin θ
∂
∂θ
(sin θ
∂f
∂θ
)+
1
ρ2 sin2 θ
∂2f
∂φ2= 2
Note. The formulae for the Laplacian:
Rectangular : ∇2f =∂2f
∂x2+∂2f
∂y2+∂2f
∂z2
Polar : ∇2f =1
r
∂
∂r
(r∂f
∂r
)+
1
r2
∂2f
∂θ2
Cylindrical : ∇2f =1
r
∂
∂r
(r∂f
∂r
)+
1
r2
∂2f
∂θ2+∂2f
∂z2
Spherical : ∇2f =1
ρ2
∂
∂ρ
(ρ2∂f
∂ρ
)+
1
ρ2 sin θ
∂
∂θ
(sin θ
∂f
∂θ
)+
1
ρ2 sin2 θ
∂2f
∂φ2
4 Linear Equations, Linear Independence and
Gram Schmidt
4.1 Assignment 4
1. Consider the system of linear equations
x+ 2y − z + w = 2
3x+ 7y − z + 2w = 6
x+ y − 3z + 5w = −4
x+ 4y + 3z + 2w = −4
(i) Find the augmented matrix.
(ii) Reduce the augmented matrix to row echelon form.
(iii) Find the general solution of the system of equations.
2. Given vectors ~a =
142
, ~b =
−105
, ~c =
35−1
, ~d =
6145
.
(i) Show that the vectors ~a,~b,~c are linearly independent.
(ii) Write ~d as a linear combination of ~a,~b,~c.
3. Determine whether the following sets of vectors are linearly independent.
(i) { [ 1, 0, 0, 1 ], [ 1, -1, 0, 0 ], [ 0, 1, 2, 1 ], [ -1, 4, 2, 2 ] }• { [ 1 0 0 1 ], [ 1 -1 0 0 ], [ 0 1 2 1 ], [ -1 1 4 3 ] }• { [ 1 -2 1 4 ], [ 0 1 -1 -1 ], [ 2 -5 3 9 ] }
(ii) { [ 1, 1, 2, 1 ], [ 1, 0, 0, 2 ], [ 4, 6, 8, 6 ], [ 0, 3, 2, 1 ] }
4. In question 3(i) and (ii), if the given set of vectors is linearly inde-pendent, use the Gram Schmidt process to create a corresponding set oforthonormal vectors.
4.2 Solutions to Assignment 4
1. (i) The augmented matrix:1 2 −1 1 23 7 −1 2 61 1 −3 5 −41 4 3 2 −4
.
(ii) Row Echelon Form: 1 2 −1 1 20 1 2 −1 00 0 0 1 −20 0 0 0 0
.
(iii) The system reduces to
w = −2
y + 2z − w = 0 =⇒ y = −2− 2z
x+ 2y − z + w = 2 =⇒ x = 8 + 5z
A general solution is given by,
S = {
xyzw
|w = −2, y = −2− 2z and x = 8 + 5z}
= {
8 +5z−2 −2z
z−2
}
= {
8−20−2
+ z
5−210
|z ∈ R}
2. (i) We use this useful result:
Proposition 1. Let A be a square matrix of size n. Then det(A) 6= 0if and only if the columns of A are linearly independent.
We can use this result to quickly check if the vectors{[1, 4, 2], [-1, 0, 5], [3, 5, -1] } are linearly independent.
det(A) = det
1 −1 34 0 52 5 −1
= 1(−25)− (−1)(−4− 10) + 3(20)
= −39 + 60 6= 0
Hence the columns of A are linearly independent.
(ii) Now to write ~d as a linear combination of ~a,~b,~c we solve the vectorequation
x~a+ y~b+ z~c = ~d
x
142
+ y
−105
+ z
35−1
=
6145
We get a corresponding system of linear equations
x− y + 3z = 6
4x + 5z = 14
2x+ 5y − z = 5
We solve this system of equations by the ususal method to get aunique solution x = 1, y = 1, z = 2.Therefore we can write
(1)
142
+ (1)
−105
+ (2)
35−1
=
6145
3. (i) S={ [ 1, 0, 0, 1 ], [ 1, -1, 0, 0 ], [ 0, 1, 2, 1 ], [ -1, 4, 2, 2 ] }We consider a linear combination of these vectors which sums to zero:
x
1001
+ y
1−100
+ z
0121
+ w
−1422
=
0000
The set of vectors are linearly independent if x = y = z = w = 0 isa unique solution!
To solve the vector equation we
– find the corresponding system of linear equations ,
– find the associated augmented matrix,
– and reduce to row-echelon form using row operations:
Here I outline the steps to reduce to row echelon form:1 1 0 −1 00 −1 1 4 00 0 2 2 01 0 1 2 0
→
1 1 0 −1 00 −1 1 4 00 0 2 2 00 −1 1 3 0
→
1 1 0 −1 00 1 −1 −4 00 0 2 2 00 −1 1 3 0
→
1 1 0 −1 00 1 −1 −4 00 0 2 2 00 0 0 −1 0
→
1 1 0 −1 00 1 −1 −4 00 0 1 1 00 0 0 −1 0
The simplified system of equations is
w = 0
z + w = 0 =⇒ z = 0
y − z + 4w = 0 =⇒ y = 0
x+ y − w = 0 =⇒ x = 0
Hence x = y = z = w = 0 is a unique solution and hence the vectorsare linearly independent.
(ii) S ={ [ 1, 1, 2, 1 ], [ 1, 0, 0, 2 ], [ 4, 6, 8, 6 ], [ 0, 3, 2, 1 ] }.We consider a linear combination of these vectors which sums to zero:
x
1121
+ y
1002
+ z
4686
+ w
0321
=
0000
To solve the vector equation we
– find the corresponding system of linear equations ,
– find the associated augmented matrix,
– and reduce to row-echelon form using row operations:
Here I outline the steps to reduce to row echelon form:1 1 4 0 01 0 6 3 02 0 8 2 01 2 6 1 0
→
1 1 4 0 00 −1 2 3 00 −2 0 2 00 1 2 1 0
→
1 1 4 0 00 1 2 1 00 −2 0 2 00 −1 2 3 0
→
1 1 4 0 00 1 2 1 00 0 4 4 00 0 4 4 0
→
1 1 4 0 00 1 2 1 00 0 1 1 00 0 0 0 0
Note that the matrix has a row of zeros which implies that we have3 equations in 4 variables and hence cannot hope to have a uniquesolution.
The simplifed system of equations is:
z + w = 0 =⇒ z = −wy + 2z + w = 0 =⇒ y = w
x+ y + 4z = 0 =⇒ x = 3w
A general solution to the system is given by
S = {
xyzw
|z = −w, y = w, x = 3w}
= {
3ww−ww
|w is a real number}
= {w
31−11
|w is a real number}
If we choose w = 1, we get x = 3, y = 1, z = −1.
(3)
1121
+ (1)
1002
+ (−1)
4686
+ (1)
0321
=
0000
This is a non-trivial linear combination that sums to zero and hencethe vectors are linearly dependent.
4. The Gram-Schmidt orthogonalisation process is defined for a set of lin-early independent vectors. Therefore we can only orthogonalize the set inquestion 3 (i).
(i) S = { [ 1, 0, 0, 1 ], [ 1, -1, 0, 0 ], [ 0, 1, 2, 1 ], [ -1, 4, 2, 2 ] }Let v1 = [1, 0, 0, 1], v2 = [1,−1, 0, 0], v3 = [0, 1, 2, 1], v4 = [−1, 4, 2, 2]
Step 1. Let
~u1 = ~v1 =
1001
~e1 =~u1
|~u1|=
1√2
1001
.
Step 2. Let
~u2 = ~v2 − Proj~u1~v2
Proj~u1~v2 =
(1,−1, 0, 0).(1, 0, 0, 1)
12 + 12
1001
=1
2
1001
~u2 =
1−100
− 1
2
1001
=
1/2−10
−1/2
~e2 =~u2
|~u2|=
√2
3
1/2−10
−1/2
.
Step 3. Let
~u3 = ~v3 − Proj~u1~v3 − Proj~u2
~v3
Proj~u1~v3 =
(0, 1, 2, 1).(1, 0, 0, 1)
12 + 12
1001
=1
2
1001
Proj~u2~v3 =
(0, 1, 2, 1).(1/2,−1, 0,−1/2)
(1/2)2 + (−1)2 + (−1/2)2
1/2−10
−1/2
= (−1)
1/2−10
−1/2
~u3 =
0121
− 1
2
1001
−
−1/2
10
1/2
=
0020
~e3 =~u3
|~u3|=
1
2
0020
.
Step 4. Let
~u4 = ~v4 − Proj~u1~v4 − Proj~u2
~v4 − Proj~u3~v4
Proj~u1~v4 =
(−1, 4, 2, 2).(1, 0, 0, 1)
12 + 12
1001
=1
2
1001
Proj~u2~v4 =
(−1, 4, 2, 2).(1/2,−1, 0,−1/2)
(1/2)2 + (−1)2 + (−1/2)2
1/2−10
−1/2
=(−11)
3
1/2−10
−1/2
Proj~u3~v4 =
(−1, 4, 2, 2).(0, 0, 2, 0)
22
0020
=
0020
~u4 =
−1422
− 1
2
1001
− (−11)
3
1/2−10
−1/2
−
0020
=
1/31/30
−1/3
~e4 =~u4
|~u4|=√
3
1/31/30
−1/3
.
5 Matrices
5.1 Assignment 5
1. Evaluate each of the following for the matrix A =
1 2 32 1 00 3 1
.
(i) A2
(ii) A3
(iii) p(A) where p(x) = −6x3 + 10x− 9.
2. Find the inverse of the matrices and verify that AA−1 = I
(i) A =
(−4 −25 5
).
(ii) A =
1 3 31 4 31 3 4
.
3. Find the eigenvalues and the eigenvectors of the given matrices.
(i)
[0 11 0
](ii)
[10 −94 −2
]
5.2 Solutions to Assignment 5
1. Evaluate each of the following for the matrix A =
1 2 32 1 00 3 1
.
(i) A2 =
5 13 64 5 66 6 1
.
(ii) A3 =
31 41 2114 31 1818 21 19
(iii) p(A) where p(x) = −6x3 + 10x− 9.
p(A) =
−185 −226 −96−64 −185 −108−108 −96 −113
.
2. Find the inverse of the matrices and verify that AA−1 = I
(i) A =
(−4 −25 5
), A−1 =
(−1/2 −1/51/2 2/5
).
(ii) A =
1 3 31 4 31 3 4
, A−1 =
7 −3 −3−1 1 0−1 0 1
.
3. Find the eigenvalues and the eigenvectors of the given matrices.
(i)
[0 11 0
]Eigenvalues: λ1 = 1, λ2 = −1,Eigenvectors: ~v1 = (1, 1), ~v2 = (1,−1).
(ii)
[10 −94 −2
]Eigenvalue:λ = 4,Eigenvectors: ~v = (3, 2)
6 Eigenvalues, Eigenvectors and Diagonaliza-
tion
6.1 Assignment 6
For the following matrices
(a) Find eigenvalues and eigenvectors
(b) Is the matrix diagonalizable? Give reasons for your answer.
(c) If it is diagonalizable, find a diagonalizing matrix P and a diagonalmatrix D, such that P−1AP = D
1.
3 1 −11 3 −1−1 −1 5
2.
0 2 22 0 22 2 0
3.
2 0 00 3 00 4 3
4.
−1 4 −2−3 4 0−3 1 3
6.2 Solutions to Assignment 6
1. Eigenvalues are {2, 3, 6} Corresponding eigenvectors are
−110
, 1
11
, −1−12
Since the matrix is symmetric of size 3, we have been able to find3 basis eigenvectors. Hence the matrix is diagonalizable.
A diagonalizing matrix P for A is obtained by taking as it’s column
vectors the 3 eigenvectors. P =
−1 1 −11 1 −10 1 2
The diagonal matrix D has the corresponding eigenvalues as it’s
diagonal elements. D =
2 0 00 3 00 0 6
2. Here again A is a symmetric matrix and hence is diagonalizable.
Eigenvalues: {−2, 4} Eigenvectors:
λ = −2 7→ X1 =
−110
, X2 =
−101
λ = 4 7→ X3 =
111
A diagonalizing matrix P for A is obtained by taking as it’s column
vectors the 3 eigenvectors. P =
−1 −1 11 0 10 1 1
The diagonal matrix D has the corresponding eigenvalues as it’s
diagonal elements. D =
−2 0 00 −2 00 0 4
3. Here A is not symmetric, so we need to see if A has three linearly
independent eigenvectors.
Eigenvalues: {2, 3}
Eigenvectors:
λ = 2 7→ X1 =
100
λ = 3 7→ X2 =
001
We have been able to find only two linearly independent eigenvec-tors. Hence A is not diagonalizable.
4. Here again A is not symmetric, so we need to see if A has sufficienteigenvectors.
Eigenvalues: {1, 2, 3}Since A has three distinct eigenvalues, it will have three linearlyindependent eigenvectors. Hence it is diagonalizable.
Corresponding eigenvectors are
111
, 2/3
11
, 1/4
3/41
A diagonalizing matrix P for A is obtained by taking as it’s column
vectors the 3 eigenvectors. P =
1 2/3 1/41 1 3/41 1 1
The diagonal matrix D has the corresponding eigenvalues as it’s
diagonal elements. D =
1 0 00 2 00 0 3
7 Operators
7.1 Assignment 7
1. Are the following operators linear or non-linear? Give a proof for youranswer.(a) Oa : x→ |x| (b) Ob : x→ ex
(c) Oc : f(x) → d2f/dx2 (d) ∇ : f(x, y, z) → (∂f∂x, ∂f
∂y, ∂f
∂z)
2. Find the expressions for the following commutator operators:
• Operating on f(x):
(a) [ d2
dx2 , x] (b) [ d3
dx3 , x2]
• Operating on f(x, y):(a) [ ∂
∂x+ ∂
∂y, xy] (b) [x ∂
∂y− y ∂
∂x, xy]
3. Let A and B be matrices defined as follows:
A =
(2 00 1
); B =
(1 00 2
).
These matrices define operators OA, OB in the following manner:
OA : x 7→ Ax
OB : x 7→ Bx
Calculate the commutator: [OA, OB]
4. Define operators acting on the wave function ψ(x, t) in the following man-ner:
• The Momentum Operator: P =~i
∂
∂x.
• The Energy Operator: E = i~∂
∂t.
• The Hamiltonian Operator: H = − ~2
2m
∂2
∂x2.
• The Position Operator: x = x.
Determine expressions for the following commutators:(a) [E, x]. (b) [x, P ].
(c) [H, P ]. (d) [H, x2].
7.2 Solutions to Assignment 7
1. Are the following operators linear or non-linear? Give a proof for youranswer.
(a) Oa : x→ |x|Oa is not linear since in general
Oa(x+ y) = |x+ y| ≤ |x|+ |y| = Oa(x) +Oa(y)
For example, | − 1 + 2| < |1|+ |2|
(b) Ob : x→ ex Ob is not linear since
Ob(x+ y) = ex+y = exey
Ob(x) +Ob(y) = ex + ey
But ex+y 6= ex + ey, as functions
For example, e1+(−1) 6= e1 + e−1
(c) Oc : f(x) → d2f/dx2
It is linear.
(d) ∇ : f(x, y, z) → (∂f∂x, ∂f
∂y, ∂f
∂z)
It is linear.
2. Find the expressions for the following commutator operators:
• Operating on f(x):
(a) [ d2
dx2 , x] = 2D
(d2
dx2◦ x)(f) =
d2
dx2(xf)
=d
dx(d
dx(xf))
=d
dx(xdf
dx+ f)
= x2d2f
dx2+df
dx+df
dx= (xD2 + 2D)(f)
(x ◦ d2
dx2)(f) = x(
d2
dx2(f))
= (xD2)(f)
[d2
dx2, x] = 2D
(b) [ d3
dx3 , x2] = 6xD2 + 6D
(d3
dx3◦ x2)(f) =
d3
dx3(x2f)
=d
dx(d
dx(d
dx(x2f)))
=d
dx(d
dx(x2 df
dx+ 2xf))
=d
dx(x2d
2f
dx2+ 2x
df
dx+ 2x
df
dx+ 2f)
= x2d3f
dx3+ 2x
d2f
dx2+ 4x
d2f
dx2+ 4
df
dx+ 2
df
dx= (x2D3 + 6xD2 + 6D)(f)
(x2 ◦ d3
dx3)(f) = x2(
d3
dx3(f))
= (x2D3)(f)
[d3
dx3, x2] = 6xD2 + 6D
• Operating on f(x, y):
(c) [ ∂∂x
+ ∂∂y, xy] = x+ y
(∂
∂x+
∂
∂y◦ xy)(f) = (
∂
∂x+
∂
∂y)(xyf)
=∂
∂x(xyf) +
∂
∂y(xyf)
= xy∂f
∂x+ yf + xy
∂f
∂y+ xf
= (xyDx + y + xyDy + x)(f)
(xy ◦ ∂
∂x+
∂
∂y)(f) = xy(
∂f
∂x+∂f
∂y)
= xy∂f
∂x+ xy
∂f
∂y
= (xyDx + xyDy)(f)
[∂
∂x+
∂
∂y, xy] = x+ y
(d) [x ∂∂y− y ∂
∂x, xy] = x2 − y2
(x∂
∂y− y
∂
∂x◦ xy)(f) = x(
∂
∂y− y
∂
∂x)(xyf)
= x2y∂f
∂y(xyf) + x2f − xy2∂f
∂x− y2f
= (x2yDy + x2 − xy2Dx − y2)(f)
(xy ◦ x ∂∂y
− y∂
∂x)(f) = xy(x
∂f
∂y− y
∂f
∂x)
= x2y∂f
∂y− xy2∂f
∂x
= (x2yDy − xy2Dx)(f)
[x∂
∂y− y
∂
∂x, xy] = x2 − y2
3. Let A,B be matrices defined as follows:
A =
(2 00 1
); B =
(1 00 2
)
These matrices define operators OA, OB in the following manner:
OA : x 7→ Ax
OB : x 7→ Bx
Calculate the commutators:
(a) [OA, OB] =
(0 00 0
)4. Define operators acting on the wave function ψ(x, t) in the following man-
ner:
• The Momentum Operator: P =~i
∂
∂x.
• The Energy Operator: E = i~∂
∂t.
• The Hamiltonian Operator: H = − ~2
2m
∂2
∂x2.
• The Position Operator: x = x.
Determine expressions for the following commutators:
(a) [E, x]. (b) [x, P ].
(c) [H, P ]. (d) [H, x2].
(a) [E, x]
[E, x] = i~∂
∂t◦ x− x ◦ i~ ∂
∂t
= xi~∂
∂t− xi~
∂
∂t= 0
(b) [x, P ]
[x, P ] = x ◦ ~i
∂
∂x− ~i
∂
∂x◦ x
= x~i
∂
∂x− [x
~i
∂
∂x+
~i]
= i~
(c) [H, P ]
[H, P ] = − ~2
2m
∂2
∂x2◦ ~i
∂
∂x− ~i
∂
∂x◦ −(~2)
2m
∂2
∂x2
= − ~3
2im
∂2
∂x2(∂
∂x) +
~3
2im
∂
∂x(∂2
∂x2)
= − ~3
2im
∂3
∂x3+
~3
2im
∂3
∂x3
= 0
(d) [H, x2]
[H, x2] = H ◦ x2 − x2 ◦ H
(H ◦ x2)(ψ(x, t)) = (− ~2
2m
∂2
∂x2◦ x2)(ψ)
= − ~2
2m
∂2
∂x2(x2ψ)
= − ~2
2m
∂
∂x(∂
∂x(x2ψ))
= − ~2
2m
∂
∂x(x2∂ψ
∂x+ 2xψ)
= − ~2
2m(∂
∂x(x2∂ψ
∂x) +
∂
∂x(2xψ))
= − ~2
2m((x2∂
2ψ
∂x2+ 2x
∂ψ
∂x) + (2x
∂ψ
∂x+ 2ψ))
= − ~2
2m(x2∂
2ψ
∂x2+ 4x
∂ψ
∂x+ 2ψ)
H ◦ x2 = − ~2
2m(x2 ∂
2
∂x2+ 4x
∂
∂x+ 2)
(x2 ◦ H)ψ(x, t) = (x2 ◦ (−~2)
2m
∂2
∂x2)(ψ)
=(−~2x2)
2m
∂2ψ
∂x2
x2 ◦ H =(−~2x2)
2m
∂2
∂x2
[H, x2] = − ~2
2m(4x
∂
∂x+ 2)
8 Fourier Series
8.1 Assignment 8
1. For the given periodic functions,
(a) Sketch the graph of the function for 3 periods.
(b) Calculate the Fourier coefficients
(c) Find the Fourier Series.
(i) A function f(x) with period T = 2π defined by
f(x) =
{0, if −π ≤ x < 0
π − x, if 0 ≤ x < π
(ii) A function f(x) with period T = 2 defined by
f(x) = |x|, −1 ≤ x ≤ 1.
(iii) A function f(x) with period T = 2π defined by
f(x) =
{−1, if −π ≤ x < 0
1, if 0 ≤ x < π
2. *
(i) Use Parseval’s result with the series in Q.2.(ii) to show that,
1 +1
34+
1
54+
1
74+ ... =
π4
96.
(ii) Use Parseval’s result with the series in Q.2.(iii) to show that,
1 +1
9+
1
25+
1
49+ ... =
π2
8.
8.2 Solutions to Assignment 8
1. For the given periodic functions,
(a) Sketch the graph of the function for 3 periods.
(b) Calculate the Fourier coefficients
(c) Write down the Fourier Series
(i) A function f(x) with period T = 2π defined by
f(x) =
{0 if π ≤ x ≤ 0
π − x if 0 ≤ x ≤ π
Solution. The function f(x) is neither even nor odd.
a0 =π
2
an =
{2
n2π, if n is odd
0, if n is even
bn =1
n
f(x) =π
4+
2
π(cosx+
1
9cos 3x+
1
25cos 5x+ ...)
+ (sinx+sin 2x
2+
sin 3x
3+ ...)
(ii) A function f(x) with period T = 2 defined by
f(x) = |x|, −1 ≤ x ≤ 1.
Solution. The function f(x) is even, since f(−x) = | − x| = |x| =f(x). Hence bn = 0.
a0 = 1
an =
{− 4
n2π2 , if n is odd0, if n is even
bn = 0
f(x) =1
2− 4
π2(cosx+
1
9cos 3x+
1
25cos 5x+ ...)
(iii) A function f(x) with period T = 2π defined by
f(x) =
{−1 if −π ≤ x < 0
1 if 0 ≤ x < π
Solution. The function f(x) is odd since f(−x) = −f(x).Hence an = 0.Therefore the Fourier coefficients are:
a0 = 0
an = 0
bn =
{4
nπ, if n is odd
0, if n is even
f(x) =4
π(sinx+
1
3sin 3x+
1
5sin 5x+ ...)
2. Parseval’s identity is:
1
L
∫ L
−L
(f(x))2dx =1
2a2
0 +∞∑
n=1
(a2n + b2n).
(i) The Fourier series for the function is
1
2− 4
π2(cosx+
1
9cos 3x+
1
25cos 5x+ ...)
Parseval’s result then gives,
1
π
∫ 1
−1
|x|2dx =1
2(1)2 +
∞∑n=1
a2n
1
π
∫ 1
−1
x2dx =1
2+
16
π4(1 +
1
34+
1
54+
1
74+ ...)
2
3=
1
2+
16
π4(1 +
1
34+
1
54+
1
74+ ...)
Rearranging
π4
96= 1 +
1
34+
1
54+
1
74+ ...
(ii) The Fourier series for the function is
4
π(sinx+
1
3sin 3x+
1
5sin 5x+ ...)
Parseval’s result then gives,
1
π
∫ π
−π
(f(x))2dx =∞∑
n=1
b2n
1
π
∫ π
−π
1dx =16
π2(1 +
1
9+
1
25+
1
49+ ...)
2 =16
π2(1 +
1
9+
1
25+
1
49+ ...)
Rearranging
π2
8= 1 +
1
9+
1
25+
1
49+ ...
