Chapter 1: Geometry of the Euclidean space

Question 1

R ⁿ is

Answer 1

the set of all n-tuples of real numbers,
Rⁿ =
{(x_1,…, x_n) : x_1, . . , x_n∈R}.

components are coordinates

Question 2

natural dot product/inner product/scalar
product

Answer 2

x · y = Σ_i=1 ,n x_iy_i,

where x = (x_1, … , x_n), y = (y_1,. .., y_n) ∈ Rⁿ

Properties:
(i) (α₁x₁ + α₂x₂) · y
= α₁(x₁ · y) + α₂(x₂ · y)
for any x₁, x₂, and y ∈ Rⁿ, and any real numbers α₁, α₂ ∈ R; distributive linear in the first var

(ii) x·y = y·x for any x and y ∈ Rⁿ symmetry

(iii) x·x > 0 for any x ∈ Rⁿ, and
x · x = 0 if and only if x = 0.
positively homogeneous

properties reflect the structure in Rⁿ

Question 3

Proposition I.1
Cauchy-Schwarz inequality

Answer 3

For any vectors x, y ∈ Rⁿ the following inequality holds:
|x · y| ≤
|x| |y|,

(where
|x| =√{x · x}, and
|y| =√{y · y} are lengths of vectors in Rⁿ)

Equality occurs IFF x = 0 or y = λx for some λ ∈ R.

Question 4

Cauchy-Schwarz inequality alt ways of writing

Answer 4

|x · y| ≤
|x| |y|,
from i=1 to n
Σxᵢ · yᵢ ≤ (Σxᵢ² Σyᵢ² )¹/²

when n=1

x₁· y₁ ≤(x₁² · y₁² )¹/² = x₁· y₁ equality
abs value= length here
|x + y|≤|x|+|y|.

Question 5

Proposition I.1
Cauchy-Schwarz inequality
PROOF L5

Answer 5

x=0 TRIVIAL CASE holds

|x| =√{x · x}
|λx-y|=√{λx-y · λx-y}
|λx-y|²={λx-y · λx-y}
assume x≠0
Consider polynomial p(λ)
= |λx-y|²
now using the properties of dot prod:
linearity, symmetry
=|λ²x² +y² -2λxy|
= λ²|x|² +|y|² -2λxy≥0
parabola in λ with at most one root
thus DISCRIMINANT ≤0
iff
(-2xy)² -4|x|²|y|² ≤0
4(xy)²-|x|²|y|² ≤0

thus inequality proved

Question 6

Proposition I.1
Cauchy-Schwarz inequality
PROOF L5 CASE OF EQUALITY

Answer 6

Looking at the case of equality
IF CASE x=0 or y =λx (colinear)
|0|≤0
|λx * λx|= λ²|x*x|= λ²|x| |x|

only if case
suppose that |xy|≤ |x||y| for some x and y
if x≠0 then we need to show that y=λx
Considering the poly: then discriminant =0
thus only one real root and hence
λ₀ that is
0=P(λ₀ ) = (λ₀x -y)² length of vector is 0 by non-neg property of dot prod
iff
λ₀x -y=0
iff
y=λ₀x

Question 7

L5 remark cauchy schwarz

Answer 7

Note that the argument used in the proof above is rather general and works for any (not
necessarily finite-dimensional) vector space V equipped with a scalar product <.,.>

Question 8

L5:
scalar product <·, ·>

Answer 8

bilinear map V × V → R that satisfies the following properties, which mimic the properties of the dot product:

(i) <α_1u_1 +α2_u_2, v> = α_1<u_1, v>+α_2<u_2, v_i> for all u_1, u_2, and v ∈ V , and all reals α_1, α_2 ∈ R; LINEAR IN FIRST ARGUMENT
(ii) <u, v> = <v, u> for all u and v ∈ V ; SYMMETRY
(iii) <u, u> > 0 for any u ∈ V , and <u, u> = 0 if and only if u = 0
NON NEGATIVITY

Question 9

Example:
Let V=C[a,b] be the space formed by all continuous vector functions f:[a,b] to R^n
is this a vector space

Answer 9

clearly, closed under linear operations (and f(x)=0?)

Question 10

Scalar product on C[a,b]
<f,g>

Answer 10

<f,g> =
∫ₐᵇ f(t)g(t) .dt

Here our vector space is infinite dimension we can show it satisfies properties of the scalar properties

Question 11

Lemma I.2 (Integral Cauchy-Schwarz inequality)

Answer 11

For any continuous vector functions f,
g : [a, b] → R^n the following inequality holds:

|∫ₐᵇ (f.g)(t) .dt| ≤
(∫ₐᵇ |f(t)|² .dt) ¹/² (∫ₐᵇ |g(t)|² .dt) ¹/²
The equality occurs if and only if f ≡ 0 or g ≡ λf for some λ ∈ R. (identically proportional?)

|<f,g>| ≤ sqrt(<f,f> <g,g>) = |f||g|

Question 12

Proof: Lemma I.2 (Integral Cauchy-Schwarz inequality)

Answer 12

Let V be a vector space of all continuous vector-functions f : [a, b] → R^n
On this vector space we consider the following scalar product
<f,g> =ᵈᵉᶠ= ∫ₐᵇ (f.g)(t) .dt where f, g ∈ V
and the function in the integral is obtained by taking the dot product of the values f(t) and g(t). It is straightforward to check that the formula above indeed defines the scalar
product on V , that is it satisfies the properties (i)-(iii) above. Now the proof of the lemma follows the same line argument as in the proof of Lemma I.1, by considering the polynomial
P(λ) = <λf − g, λf − g>.

Question 13

Corollary I.2 (Triangle inequality). Of CS

Answer 13

For any vectors x, y ∈ Rⁿ the following inequality holds:
|x + y|≤|x|+|y|.

Equality occurs IFF x = 0 or y = λx for some λ> 0 in R

(proportional/colinear for equality)

Question 14

Corollary I.2 (Triangle inequality consequence of C-S).
PROOF

For any vectors x, y ∈ Rⁿ the following inequality holds:
|x + y|≤|x|+|y|.

Equality occurs IFF x = 0 or y = λx for some λ> 0 in R

Answer 14

proof:
By linearity and symmetry defn using values and lengths
|x + y|²
= (x + y) · (x + y)
= |x|² + x · y + y · x + |y|²
= |x|² + 2x · y + |y|²
≤|x|² + 2 |x · y| + |y|²

≤|x|² + 2 |x| |y| + |y|²
= (|x| + |y|)²

(second inequality used the Cauchy-Schwarz inequality)

Equality in the triangle inequality implies equality in the Cauchy-Schwarz inequality, and hence, implies that x = 0 or y = λx for some λ > 0.
shown in lecture
(if x=0 |y|²=|y|², if y= λx LHS:
|x + λx|= |(1+λ)x|= sqrt((1+λ)x·(1+λ)x )
RHS: |x|+ |λx| = |1+λ| |x| shows equality)

Conversely, if x = 0 or y = λx, where λ > 0, then the triangle inequality becomes an equality.

Question 15

Suppose |x+y|² =( |x|+|y|)²

Use modulus and dot product relationship

Answer 15

(x+y) · (x+y)
= ( |x|+|y|)²
by cauchy schwarz conclude x=0 or y=λx

x not eq 0 (1+λ)² |x|² = (1-λ)²) |x|² if λ<0
however plugging y= λx into inequality |x+y| <= |x| + |y| we rule out the case when λ<0

Question 16

distances in R^n

Answer 16

dist(x,y)
=|x−y|= (x−y)·(x−y)

From MATH2051 we also know that the dot product allows us to compute lengths of curves in R^n

Question 17

Distances in R^n

Answer 17

dist(x,y)=|x−y|= (x−y)·(x−y).

From MATH2051 we also know that the dot product allows us to compute lengths of curves in R^n

Question 18

Triangle inequality in form for distances

Answer 18

In particular, the triangle inequality can be written in the following form
for any vectors x, y, and z ∈ Rn. (Make sure that you can explain why.)

dist(x, y) less than it equal to
dist(x, z) + dist(z, y)

Question 19

|x-y| inequality for triangle

Answer 19

less than or equal to |x-z| + |z-y|

iff
|u+v| <= |u| +|v|
u=x-z and v= z-y

Question 20

x·y./ (|x| |y|)

Answer 20

x·y./ (|x| |y|) in [-1,1]
hence choose cos theta

also choosing thete in [0, pi]
assuming x,y non zero vectors

Question 21

to compute the cosine of angles
θ

between non-zero vectors x and y

Answer 21

by the formula
cosθ= x·y./ (|x| |y|)

RHS cos θ for some θ is a consequence of the Cauchy-Schwarz inequality: it guarantees that the quotient on the right hand-side above takes values in the interval [−1, 1].

The equality case in the Cauchy-Schwarz inequality says that the angle θ between non-zero vectors x and y equals πk, where k ∈ Z, if and only if the vectors are collinear.

Recall that vectors x and y are called orthogonal if x · y = 0, i.e. the angle between them equals π/2 + πk, k ∈ Z.

Question 22

Pythagorean theorem:

Answer 22

vectors x and y are orthogonal if and only if |x + y|^2 = |x|^2 + |y|^2.

Particular case of
Distance using Cauchy Schwartz

Question 23

. to compute volumes of parallelotopes (not covered in lecture but was in exercise !)

Answer 23

in more detail, for a system e1, . . . , ek of k linearly independent vectors the k-dimensional parallelotope Pk, spanned by them, is defined as
Pk ={t1e1 +…+tkek :ti ∈[0,1]}. Its k-dimensional volume is given by the formula

Vol_k (P_k) =
√[
Det(
[ e ₁. e₁ e ₁.e₂ …. e₁.eₖ]
…..
[ eₖ. e₁ eₖ. e₂ …. eₖ.eₖ])

In particular, if the ei’s are pair-wise orthogonal (that is Pk is an orthotope ), then we obtain
Volk(P_k)= SQRT((e1 ·e1)···(ek ·ek) )=|e_1|…..|e_k|

Question 24

def 1.2 vector subspace’

linear vector space

Answer 24

A subset X ⊂ R^n st for any x, y ∈ X and any a, b ∈ R we have ax + by ∈ X.

closed under linear combos,
subspace commonly solutions to linear equations
0 is always an element of a vector subspace

we use properties of dot prod and vector space properties

we define a subspace as a span of a collection of vectors min # give basis

Question 25

Which ones are vector subspaces?

Example I.3. Let X₁, X₂, X₃ ⊂ R² be subsets defined as:

X₁ = {(x₁, x₂) ∈ R²: x₁ = x₂},

X₂ = {(x₁, x₂) ∈ R²: x₁= x₂ + 1},

X₃ = {(x₁, x₂)∈ R² : x₁=x₂²}

Answer 25

X1 is a vector subspace, but X2 and X3 are not.

1) check that if x = (x₁, x₂)
and y = (y₁, y₂) are vectors from X1, then any linear combination ax + by also lies in X1.
ax + by = a(x₁, x₂) + b(y₁, y₂)
= (ax₁+ by₁ , ax₂ + by₂)
Since x₁= x₂ and y₁=y₂ we conclude that ax₁ + by₁ = ax₂ + by₂ , that is ax + by lies in X1.

2) The set X2 is not a vector space, since it does not contain the zero vector 0 = (0, 0); by Definition I.2 any vector space should do so.

3) X3 is not a vector space, note that the vectors (1, 1) and (1, −1) lie in X3, but the sum (2, 0) = (1, 1) + (1, −1) does not.

Question 26

orthogonal

Answer 26

when vectors are orthogonal angle is (pi/2)k
ie x.y=0

Question 27

pythagorean thm

Answer 27

applies to sides which are orthogonal

IFF |x+y|^2 = |x|^2+|y|^2

Question 28

def 1.4 linearly dependent

Answer 28

A collection of non-zero vectors x₁, . . . , xₖ ∈ Rⁿ is called linearly dependent if there
exists a collection of real numbers α₁, . . . , αₖ ∈ R such that not all of the αᵢ’s are equal to zero and
α₁x₁ + α₂x₂ + · · · + αₖxₖ = 0.

A collection of vectors that are not linearly dependent is called linearly independent.

Question 29

Self-Check Question I.4. Let x₁, . . . , xₖ ∈ Rⁿ be a collection of non-zero vectors that are pair-wise orthogonal, that is
x_i·x_j = 0 for all i not equal to j.

Can you show that these vectors are linearly independent?

Answer 29

suppose linearly dependent
consider
a₁x₁+ . . . +aₖxₖ=0
we look for these to be non zero reals
Taking x_i * (a₁x₁+ . . . +aₖxₖ)=0
gives only non zero dot prod as
a_i (x_i x_i)=0
we have x_ix_i not equal to 0 as these are non zero vectors and |x_i|^2 = x_i*x_i
thus must have a_i=0
repeat argument for all a_i
thus linearly independent

Question 30

x,y in R linearly dependent if

Answer 30

2 vectors linearly dependent IFF x=0 or y=kx k in R (covector)
scalar multiples a_1x+a_2y=0 s.t a_1 not equal to 0 x=(a_1/a_2)y=0

Question 31

spans

Answer 31

Let x₁, . . . , xₖ ∈ Rⁿ be a collection of non-zero vectors
collection spans a vector subspace X ⊂ R^n if for any x ∈ X there exist k real numbers αi ∈ R, where i = 1, . . . , k, such that
x = α₁x₁ + α₂x₂ + · · · + αₖxₖ

Question 32

Let x₁, . . . , xₖ ∈ Rⁿ be a collection of non-zero vectors that lie in a vector subspace X ⊂ R^n. Then EQUIVALENT STATEMENTS

Answer 32

(i) x1, . . . , xk is a maximal system of linearly independent vectors in X;
(ii) x1, . . . , xk is a minimal system of vectors that span X;
(iii) x1, . . . , xk is a system of linearly independent vectors that span X;
(iv) for any x ∈ X there exist a unique collection of real numbers α1, . . . , αk such that
x = α1x1 + α2x2 + · · · + αkxk.

Question 33

def 1.5 BASIS

Answer 33

Let X ⊂ R^n be a vector subspace. A collection of non-zero vectors x1, . . . , xk ∈ X that satisfies one of the equivalent statements (i)-(iv) in Proposition I.3 is called a basis of X.

a basis exists of at most n vectors in R^n
we prefer an orthonormal basis

Question 34

prefer an orthonormal basis

Answer 34

If its a basis then any vector x in V can be expressed as a unique linear combo
x= a_1x_1 +…+a_kx_k for a_i in R
if basis is orthonormal then coeffs are easier to compute

a_i = (x*x_i) bc dot prods =1 or =0 for that i’th component

Question 35

DIMENSION of a vector space X

Answer 35

BASIS VECTORS
f x_1, . . . , x_k and y1, . . . , ym are two different basis of X, then k = m,
the number of vectors in a basis of a fixed vector space X is always the same.
This integer is the dimension of a vector space X.

Question 36

E.G Let X,Y be 2 vector spaces in R^n

Show that if X⊆ Z and dim Z= dim X
then X=Z

Answer 36

The dimensions mean we can describe using a basis with the same number of vectors
Let x∈X then
x=a_1x_1+….+a_kx_k for some reals basis {x_1,…,x_k}
X⊆Z means x∈Z so
x=b_1z_1+….+b_kz_k for some reals basis {z_1,..z_k}
that means
the basis for Z is also a basis for X, as this is true for any x in X
take z in Z,
we use the basis
z=c_1z_1+….+c_kz_k
as this is a basis in X also, z in X
so Z⊆X
so X=Z

Question 37

ONE DIMENSIONAL SUBSPACE

Answer 37

in R^n
ℓ_v={tv: t∈R }
v ∈R^n fixed non zero DIRECTION vector
Lines through O
basis 1 vector

Question 38

Plane through 0∈R^n

Answer 38

vector subspace dim 2

Question 39

vector subspace dim 2

Answer 39

Plane through 0∈R^n
ℓ_v ⊥
{(x,y,0,…,0} x,y ∈R^n}

Question 40

L5: theory existence of orthonormal basis

DEFN ORTHOGONAL collection

Answer 40

Existence of an orthonormal basis. Let V⊂R^n be a subspace.

A collection of vectors x₁,…,xₖ ∈ Rⁿ is called ORTHOGONAL if vectors are pairwise orthogonal
xᵢ*xⱼ=0 for all i not equal to j

Note that if a collection of vectors is orthogonal, then the vectors are linearly independent.

Question 41

ORTHONORMAL collection

Answer 41

A collection of vectors x₁,…,xₖ ∈ Rⁿ is called ORTHONORMAL if
xᵢ*xⱼ=
{1 if i=j
{0 if i not equal to j

PAIRWISE ORTHOGONAL
UNIT LENGTH

Question 42

L5:
EXISTENCE OF A BASIS
Theorem I.3.

there MAYBE a question on the exam about this :)

Answer 42

Let V ⊂ Rⁿ be a vector subspace. Then there exists an orthonormal basis in V .

Question 43

ORTHOGONAL COMPLEMENT OF X

Answer 43

X⊥
={y∈R^n y*x =0 ∀x∈X}⊂R^n

vector subspace X
all vectors orthogonal to all x∈X

Question 44

dimensions of X and X⊥ in R^n

Answer 44

complimentary
dim X=k
dim X⊥=n-k

Question 45

For ℓ_v through O in R^n
ℓ_v ⊥ is

Answer 45

a PLANE ORTHOGONAL to ℓ_v

Question 46

is
X⊥ a VECTOR SUBSPACE?

Answer 46

yes
0 in X⊥
for x,u,v in X⊥
(au+bv)x= aux+bv*x=0 closed
(checking orthogonal to x in X checks its in our set )

Question 47

V=span(e_1.e_2)
then V⊥

Answer 47

is a line formed by e_3 axis
orthogonal to e_1 and e_2
thus orthogonal to all combos of these (the plane)

Question 48

(X⊥)⊥

Answer 48

=X
complementary dims
y in (X⊥)⊥ by definition is orthogonal to all vectors orthogonal to all in X
dim(X)=k
dim(X⊥)=n-k
thus
dim(X⊥)⊥ = n-(n-k)=k

x in X is not in X⊥ but is orthogonal to all y in X⊥

Question 49

LEMMMA I.4
two lines coincide as sets IFF

Answer 49

Two lines
ℓ_{p,v}
ℓ_{q,w}
coincide as sets
ℓ_{p,v}=ℓ_{q,w}

IFF
the direction vectors v,w are linearly dependent and p-q=t_o v for some t_0 in R

Question 50

PROOF
Two lines
ℓ_{p,v}
ℓ_{q,w}
coincide as sets
ℓ_{p,v}=ℓ_{q,w}

IFF
the direction vectors v,w are linearly dependent and p-q=t_o v for some t_0 in R

Answer 50

suppose lines coincide
q∈ℓ_{p,v}
hence there exists t_0∈R s.t
q= t_0 v +p
also as lines coincide
q+w∈ℓ_{p,v}
there exists t s.t
q+w=tv+p
ℓ_{q,w}hence
(t-t_0)v=w
thus v&w linearly dependent

conversely if
q=p+t_0v for some t_0 in R
then q∈ℓ_{p,v}
since the non zero vectors v and w are linearly dependent we can assume
w=av for non zero a
concluding

ℓ_{q,w}=
{q + sw : s ∈ R} =
{p + (t_0 + sα)v : s ∈ R} =
=ℓ_{p,v}

Question 51

Proposition I.5.

The lines, viewed as subsets of R^2 of form
ℓ_{p,v}
={p+tv: t ∈ R} ⊂ R^2

satisfy Euclid’s axioms A1-A5.

Question 52

PROOF

Question 53

EUCLIDS AXIOMS

Answer 53

A1. For any two points there exists a line that goes through both of them.

A2. There exists at most one line that passes through two distinct points.
A3. Every line contains at least two distinct points.
A4. There exist three points that do not lie on a straight line.
A5. (Parallel axiom). Let ` be a line and p a point that does not lie on . Then there exists a unique line that contains p and does not intersect the line .

Question 54

LINEAR MAP/OPERATOR
defn

Answer 54

A map L : V → W is called linear if
L(α_1u_1 + α_2u_2) =
α_1L(u_1) + α_2L(u_2)
for all α_1, α_2 ∈ R, u_1, u_2 ∈ V.

Question 55

isomorphism

Answer 55

A linear map L : V → W is called isomorphism if it is bijective, that is both
injective and surjective.

we can check these looking at rank and nullity

Question 56

C approx R^2

Answer 56

Euclidean plane as the set of complex X span {1,i} dim 2
isomorphic to R^2
Z to (Re(z), im(z)) bijective linear map

Question 57

PROJECTION ONTO LINE
Is it a linear map

Answer 57

orthogonal projection on ℓ_v
={tv, t in R}
For line through 0 with |v|=1
map
L:R^n to ℓ_v s.t
L(u) =(u * v)v

L is a linear map:
L(au_1+bu_2) using dot prod properties
=
((au_1+bu_2)* v
=
a**u_1 * v ** + b u_2 * v
= aL(u_1) +bL(u_2)

for any w in ℓ_v L(W)=w

Question 58

PROJECTION ONTO LINE PROPERTY
for any w in ℓ_v
L(w)

What happens to a w in line
V is unit vector

Answer 58

=(w * v) * v
=(t |v|^2 ) * v (tv * v ) *v
(|v|^2 =1 as unit vector)
=tv =w

Question 59

PROJECTION ONTO LINE
PROPERTY
image of L

Answer 59

coincides with ℓ_v
rank L = dim (ℓ_v)= 1

Question 60

PROJECTION ONTO LINE
PROPERTY
Ker L

Answer 60

={v in R^n| L(u)=0}
={v in R^n : u * v = 0}
=
ℓ_v⊥
as orthogonal to v

Question 61

ORTHOGONAL PROJECTION ONTO X
Π
linear map

Answer 61

this is a linear map
For orthonormal basis
u_1,…,u_k for X

Π:R^n to X
u to v = sum _{i=1,k} (u * u_i) u_i

diagram: vector projected onto another
same direction as projected onto vector
different length

im Π= X
ker Π=X⊥

Question 62

KERNEL for linear map
L:U to V

Answer 62

KerL =
{u ∈ R^n: L(u) = 0} ⊂V
PREIMAGE of 0
vectors onto 0

Nullity is the dimension

Question 63

KERNAL for PROJECTION INTO LINE
map L

Answer 63

{u ∈ R^n: u · v = 0} = `
ℓ_v⊥

Question 64

IMAGE

Answer 64

{w in W for which there there exists u in V s.t L(u)=w}⊂ W

RANK is the dimension

Question 65

IMAGE
for PROJECTION INTO LINE

Answer 65

linear map
For orthonormal basis
u_1,…,u_k for X

Π:R^n to X
u to v = sum _{i=1,k} (u * u_i) u_i

diagram: vector projected onto another
same direction as projected onto vector
different length

im Π= X

Question 66

prop 1.6
A linear map is INJECTIVE

Answer 66

A linear map L : V → W is
INJECTIVE
IFF
KERNAL TRIVIAL
nul L =0

Question 67

PROPN 1.6
A linear map is SURJECTIVE

Answer 67

IFF
rank L = dim W

Question 68

PROOF
A linear map L : V → W is
INJECTIVE
IFF
KERNAL TRIVIAL
nul L =0

Answer 68

suppose the contrary
there exists v non zero s.t L(v)=0 but injectivity contradicts as L(0)=0

converse
kernel trivial
suppose contrary, there exists v_1,v_2 in V s.t
L(v_1)=L(v_2) then by linearity
L(v_1-v_2)=L(v_1)-L(v_2)=0
but kernel trivial contradicted as v_1-v_2 ∈kernel

Question 69

PROOF
A linear map L : V → W is
SURJECTIVE

IFF
rank L = dim W

Answer 69

suppose surjective then for all w in W there exists v in V s.t L(u)=w

thus dim W=rank L

converse
suppose tank L=dim W
then dim ({w in W:L(u)=w for u in V})
then for all w in W there exists u in U s,t L(u)=w

Question 70

PROJECTION ONTO LINE
surjective?
rank and kernel

Answer 70

for any w = tv ∈ ℓ_v, we have
L(w) = (w · v)v =
t|v|^2v = tv = w,
where we used that |v| = 1.

Thus, the image of L coincides with ℓ_v, and hence, rank L = dim ℓ_v = 1.

KerL =
{u ∈ R^n: L(u) = 0}
= {u ∈ R^n: u · v = 0}
= ℓ_v⊥

where we used that v =\ 0 in the second equality, and Definition I.7 in the last.
nulL = dim Ker L = dim ℓ_v⊥

surjective IFF rank L= dim W so map L: R^n to ℓ_v surjective

Question 71

ROTATION angle θ around axis
is it a linear map

Answer 71

LINEAR MAP
R_θ(z)= exp(iθ) z

R_θ(a_1z_1 +a_2 z_2)=
exp(iθ) (a_1z_1 +a_2 z_2)
= a_1R_θ(z_1)+a_2 R_θ(z_2)

commutative and distributive
rotations preserve length and angles of vectors
another orthonormal basis is Rθ(e_1), Rθ(e_2)

Question 72

ROTATION angle θ around axis
formula

Answer 72

Rθ(v)=
(Rθ(v)* Rθ(e_1)Rθ(e_1) + (Rθ(v)* Rθ(e_2))Rθ(e_2)

(Rθ(v)* Rθ(e_1)=
|(Rθ(v)||Rθ(e_1)||cosφ
= |v||e_1| cosφ
= v*e_1

Question 73

REFLECTION THROUGH H
LINEAR MAP

Answer 73

H⊂R^n
H is HYPERPLANE vector subspace dim H =n-1
Let v in R^n be a vector orthogonal to H
v*y =0 for all y in H

R_H: R^n to R^n
R_H(V)=-v for vectors orthogonal
R_H(y)=y for all y in H

vectors in H invariant,
vectors orthogonal change

Question 74

example
R_{e_1)(z)= z conjugate

Answer 74

conjugate only i sign changes
so its a rotation Kernal dim 0
rank L=n-0=n

reflection in Re z axis

in euclidean plane, rank of rotation =2
surjective and injective

Question 75

PROPN
RANK-NULLITY THM

Answer 75

Let L : V → W be a linear map. Then
dim V = rankL + nulL

In particular, rankL <= dim V

Question 76

for any linear map
rankL ≤

Answer 76

rank L ≤min{dimV, dim W}

Question 77

How to represent v using standard basis e_1 and e_2

Answer 77

v =
(v * e_1) e_1 + (v*e_2)e_2
using standard basis in R^2

Question 78

corollary of rank nullity

Answer 78

Let L:V to W be a linear operator then
(i) if dim ≤ dim W, the operator L has maximal rank if and only if it is injective; Nul L=0

(ii) if dim V ≥ dim W, the operator L has maximal rank if and only if it is surjective. Rank = dim W

if dim V = dim W, then a linear operator L has maximal rank IFF if it is an isomorphism

proved
i)rank L= dim V IFF nul L =0 IFF injective from prev propn

Question 79

example: applying Rank nullity thm to orthogonal complement/PROJECTION
Π:R^n to X

Answer 79

dim X= dim (Im Π) = rank Π =k
dim X⊥=dim ker Π = nul Π =n-k

R^n = X union X⊥
iff
X intersection X⊥ ={0}
iff
v in R^n decomposed as
x=x+y with x in X y in Y
x projected is Π(x)

Question 80

example
V be a space of m × n matrices
W R^{m×n}.

Consider a basis of V:

Answer 80

basis of matrices eji (components 0 except ith row jth col)

let e_l standard basis in W: (lth component 1 , all 0)

Consider the linear operator
I_{m,n} : V → W
sends a matrix eji to the vector e_{n(j−1)+i}
j = 1, . . . , m and i = 1, . . . , n.
the linear operator is a linear isomorphism

Question 81

C[a,b]

Answer 81

continuous dunctions f:[a,b] to R^n
is a vector space
f(x)=0 is continuous
closed as af+bg is also a continuous funct

Question 82

Let
{v_1,..,v_n} and{w_1,..,1_n} be bases in V and W
then
any linear operator L:V to W can assign m x n matrixA_L

Answer 82

components
(a_ji)

L(v_i) =
sum_{j=1,m} a_{ji}w_j
i=1,..,n

Question 83

A_l used for any linear operator has properties

Answer 83

L to A_L
(i) A_L=0 if and only if L = 0,
(ii) Aλ_L = λA_L for any λ ∈ R and L : V → W,
(iii) A_{L+S} = A_L + A_S for any linear operators L, S : V → W,

(iv) A_{L◦S} = A_LA_S for any linear operators S : Z → V and L : V → W.
(v) Ker L = {sum x_iv_i}, where (x1, . . . , xn) is a solution to the linear system
sum_{i=1,n}a_{ji}x_i = 0 for any j = 1, . . . , m;
if n = m, then nul L = 0 if and only if det /= 0.(square matrix)

(vi) rank L = rank A_L, rank of the matrix AL; if n = m, we see that L has maximal rank if and only if det A_L /= 0.

Question 84

when is L an isomorphism
L:R^n to R^n

Answer 84

nulL = 0 ⇔ rankL = n ⇔ det A_L /= 0 ⇔ L is an isomorphism

if L satisfies these its invertible and the inverse L-1 also satisfies

LINEAR ISOMORPHISMS ~ INVERTIBLE LINEAR OPERATORS

Question 85

if L and S are two maps isomorphic then

Answer 85

their composition is

Question 86

General linear group GL(n)

Answer 86

The collection of all linear operators that satisfy any of the hypotheses is
called the general linear group and is denoted GL(n).

nulL = 0 ⇔ rankL = n ⇔ det A_L /= 0 ⇔ L is an isomorphism

Question 87

matrix of operator

Answer 87

A_L
DEPENDS ON CHOICE OF BASIS
[a_11 ……a_1n]
[…]
[a_m1…..a_mn]
m x n matrix

first column found from L(v_1)
last column from L(v_n) for vectors in basis

Question 88

Gram-Schmidt process.

exam q?

Answer 88

The process of transforming any basis into an orthogonal basis that was described in the proof of Theorem I.3 is called the Gram-Schmidt process

Question 89

what do I need to find A_L?

GL(n)

Answer 89

a basis {e1, . . . , en} in R^n, we may identify linear operators L with n × n matrices A_L,
GL(n) =
{n × n-matrices A : det A /= 0}.

Question 90

two matrices of a linear operator?

Answer 90

If {e¯1, . . . , e¯n} another basis, and and linear op has matrix A¯_L
then

A¯L = C⁻¹A_L C
C = (c_ji),
e_j = sum{i=1,n} c_{ji}e¯_i
.
Here C is a matrix for transition map, maps one basis to the other. In
particular, by the multiplicative property of the determinant we conclude that
det A¯L = det(C−1A_LC)
= det C−1 det AL det C = det AL,

as det C −1 = 1/ det C.

Question 91

RANK of a matrix

Answer 91

maximal # linearly indep cols of a
or
max #linearly indep rows

if A in REF then rank = #nonzero rows

Question 92

PROJECTION ONTO LINE example of matrix

Answer 92

linearly infep cols ~ rank of matrix =1

L(u) = (uv)v matrix found by
L(e_i)=(e_i v)v
= v_i V

matrix is
[v_1v, v_2v,… v_nv]

each col ~ projection of e_i onto v, scaled by v_i component v

v spanning 1d subspace: nullity = n-1

Question 93

example rotation matrix

Answer 93

Rθ(e_1) = (cosθ, sinθ)
Rθ(e_2) = (-sinθ, cosθ)
taking these as cols

A_Rθ =
[cosθ, -sinθ]
[-sinθ, cosθ]
Rθ(e_1) Rθ(e_2)

theyre linearly indep so
rank =2
nullity=0

Question 94

composition of rotations

Answer 94

is a rotation
composition
Rθ * Ra = R(θ+a)

matrices
A_Ra * A_Rθ

Question 95

ROTATIONS
preserve

Answer 95

ROTATIONS are LINEAR MAPS that PRESERVE DOT PRODUCT
L(uv) =uv for all u,v in R^n

Question 96

matrix for R(z)= z conjugate

Answer 96

R(e_1)=e_1
R(e_2)= -e_2

CONJUGATION~REFLECTION
in basis e_1
[1 0]
[0 -1]

and is therefore a rotation

Question 97

ORTHGONAL LINEAR OPERATORS

Answer 97

linear operator L : R^n → R^n is called orthogonal (or an orthogonal transformation), if it preserves the dot product:

L(u) · L(v) = u · v
for all u, v ∈ R^n

satisfy
=|L(u)||L(v)| cos φ
=u*v
PRESERVE LENGTHS

Question 98

REFLECTIONS/ROTATIONS

ORTHOGONAL LINEAR OPERATORS

Answer 98

reflections, rotations
contained in
orthogonal linear operators

contained in
euclidean isometries

Question 99

ORIENTATION PRESERVING

Answer 99

An isomorphism (invertible operator) L : R^n → R^n is called orientation preserving,
if its matrix AL with respect to some (and hence any) basis satisfies det A_L > 0. Otherwise, an
isomorphism L is called orientation reversing.

Question 100

example
reflection through coordinate hyperplane

H =
{(x_1, . . . , x_{n−1}, 0) : x_1, . . . , x_{n−1} ∈ R} ⊂ R^n

formula
matrix (standard basis)

Answer 100

M(x_1, . . . , x_{n−1}, x_n) =
(x1, . . . , x_{n−1}, −xn)

matrix
[1 0 …0]
[0 1 0…0]
[…]
[0….. 0 -1]

det =-1
concluding orientation reversing

Question 101

Write down the matrix of the reflection through the coordinate hyperplane
H˜ =
{(0, x_2, . . . , x_n) : x_2, . . . , x_n ∈ R} ⊂ R^n
.

Answer 101

M(x_1,..,x_n) = (-x_1,x_2,..,x_n)
A_m=
[-1 0 …0]
[0 1 0…0]
…
[0…0 1]

Question 102

when do we have a reflection through H

Answer 102

Let H in R^n be a vector subspace of dim n-1
A linear operator is the reflection through H

if M_A(v) =v for any v in H
M_H(v) = -v for any v orthogonal to H

Question 103

PROPN
basic properties for orthogonal operator

Answer 103

Let L : R^n → R^n be an orthogonal operator. Then:

(i) it is invertible, and the inverse operator L−1is also orthogonal;
(ii) if S : R^n → R^n is another orthogonal operator, then the composition L ◦ S is also an orthogonal operator;

(iii) the operator L preserves lengths of vectors, that is |L(u)| = |u| for any u ∈ R^n.

Question 104

proof PROPN for orthogonal op properties

Answer 104

Proof. We first check property (iii):
|L(u)|^2 = L(u) · L(u) = u · u = |u|^2
for any u ∈ R^n.

Now property (iii) implies that nul L = 0. Hence, any of the hypotheses in (I.5) is
satisfied, and we conclude that L is invertible. L⁻¹(u) · L⁻¹(v) =
L(L⁻¹(u)) · L(L⁻¹(v)) = u · v
for any u and v ∈ R^ n. Thus, property (i) is verified.

(ii) follows by the
repeated application of relation
(L ◦ S)(u) · (L ◦ S)(v) = L(S(u)) · L(S(v)) = S(u) · S(v) = u · v for any u and v ∈ R^n
Rank-nullity shows injective and surjective then isomorphism

Question 105

orthogonal group

Answer 105

collection of all orthogonal operators
{A in GL(n): A^TA=E}

L : R^n → R^n forms the orthogonal group O(n).

Question 106

EUCLIDEAN ISOMETRY

Answer 106

A (not necessarily linear!) map T : R^n → R^n is called a Euclidean isometry, if it preserves distances between points, that is
dist(u, v) = dist(T(u), T(v)) for all u, v ∈ R^n

Question 107

is a translation a euclidean isometry

Answer 107

P:R^n to R^n
u to u+p

dist(P(u),P(v))
= |P(u)-P(v)|
= |u +p -v+p|
=|u-v|
= dist(u,v)
for all u,v in R^n

Question 108

COROLLARY
euclidean isometries and orthogonal operators

Answer 108

Any orthogonal operator L : R^n → R^n is a Euclidean isometry.

Proof. Since L is a linear operator, then using property (iii) in Proposition I.9, we obtain
dist(u, v) = |u − v| = |L(u − v)| = |L(u) − L(v)| = dist(L(u), L(v))
for any u and v ∈ R^n

Question 109

Euclidean isometry such that T(0)=0

Answer 109

Theorem I.11. Let T : R^ n → R^n be a Euclidean isometry that fixes the origin, that is T(0) = 0.
Then T is an orthogonal linear operator.

Proof. For MATH5113M only

T fixes origin THEN it has to be linear

Question 110

PROOF L5

Question 111

Corollary i.12 an euclidean isometry is

Answer 111

Corollary I.12. Let T : R^n → R^ n be a Euclidean isometry. Then it is the composition of an orthogonal transformation T_0 followed by a translation P, that is T = P ◦ T_0.

Question 112

proof

Answer 112

T(0)=p trivial if p=0
p not equal to 0
p(x)=x+p bijective inverse p-1(x) =x-p

composition p^-1 T
p,T are EIs
p^-1 T(0)=0 so by thm
p-1 . T = T_0

unique composition determined by value of T
as composition of EIs is an EI

thus its an orthogonal trans uniquely determined by the value of…

Question 113

scalar product on C[a,b]

Answer 113

<f,g> =∫ₐᵇ f(t)g(t).dt

1) <a_1f_1 +a_2f_2,g> = a_1<f_1,g> +a_2<f_2,g>

2)<f,g>=<g,f>

3) <f,f> =0 iff f=0 by CS

Question 114

INTEGRAL CAUCHY SCWARTZ LEMMA

Answer 114

for any functs f,g in C[a,b]
|∫ₐᵇ f*g .dt| <=
SQRT[|∫ₐᵇ f^2 .dt||∫ₐᵇg^2.dt|]

equality IFF f=0 or g = λf for λinR (identically proportional)

|<f,g> |<= SQRT[<f,f><g,g>]

Question 115

integral Cauchy schwarz lemma
PROOF

|<f,g> |<= SQRT[<f,f><g,g>]

equality IFF f=0 or g = λf for λinR (identically proportional)

Answer 115

Let V be a vector space of all the continuous functions f:[a,b] to R^n
consider polynomial p(λ)=<λfg, λfg>

using the properties of <> mean we will have real roots discriminant gives the inequality

equality from CS lemma

Question 116

might be exam q on this
THM
when does an orthonormal basis exist

Answer 116

Let V⊂R^n be a vector subspace. Then there exists an ORTHONORMAL BASIS IN V

Question 117

PROOF

Answer 117

Let P : R^n → R^n be a translation by p = T(0). Then the inverse map P−1 is also a translation
(by −p), and an isometry. Since the composition of two isometries is an isometry, we see that the
map P−1 ◦ T is an isometry. Besides, we have P−1◦ T(0) = P−1
(p) = p − p = 0.
Thus, by Theorem I.11 the map P
−1 ◦ T is an orthogonal transformation. Setting T_0 = P−1 ◦ T, we obtain that T = P ◦ T_0.

Question 118

PROP i.13
How can we check a linear operator is orthogonal

Answer 118

Let (e_i) be an orthonormal basis in R^n.
Then a linear operator L : R^n → R^n is orthogonal
IFF
its matrix A wrt basis (e_i) satisfies A^TA = E,
(identity matrix)

Question 119

PROOF
Let (e_i) be an orthonormal basis in R^n.
Then a linear operator L : R^n → R^n is orthogonal
IFF
its matrix A wrt basis (e_i) satisfies A^TA = E,
(identity matrix)

Question 120

L is orthogonal ⇒ A^TA = E.

Answer 120

true

Question 121

e.g is this an orthogonal operator matrix
[0.5 √3/2]
[- √3/2 0.5]

Answer 121

By propn we have
A^T A =E
orthogonal operator

we can also show cols form orthogonal basis themselves

Question 122

e.g is this an orthogonal operator matrix

[0.5 √3/2]
[√3/2 0.5]

Answer 122

not
lengths of cols not equal to 1

Question 123

Corollary I.14. Let e1, e2 be an orthonormal basis in a Euclidean plane R^2. Then the matrix A of
any orthogonal operator L : R^2 → R^2 has one of the following forms

Answer 123

A_L =
[cos θ −sin θ]
[sin θ cos θ]
det>0 orientation preserving

or
A_l=
[cos θ sin θ]
[sin θ − cos θ]

det<0 orientation reversing
0 <=θ < 2π.

Question 124

PROOF
A_L =
[cos θ −sin θ]
[sin θ cos θ]
det>0 orientation preserving

or
A_l=
[cos θ sin θ]
[sin θ − cos θ]

det<0 orientation reversing
0 <=θ < 2π.

Answer 124

Suppose that the matrix A of L
[a c]
[b d]

Then relation A^TA=E gives
a^2+b^2=1
ac+bd=0
c^a +d^2=1
set
a=cosθ
b=sinθ
for some
0 <= θ < 2π

c=-tsin θ
d= tcos θ
for some t ∈ R
obtaining
t^2=1
thus t=+-1
corresponds to det sign

Question 125

summary of matrices for orthogonal operators

Answer 125

orientation preserving orthogonal operators R^2 → R^2 are precisely
rotations (through an angle θ in the anti-clockwise direction), and orientation reversing orthogonal
operators R^2 → R^ 2 are precisely rotations (through an angle θ in the clockwise direction) followed by
a reflection (mirror symmetry) in e_1

IF LINEAR OP
matrix cols

L(v)=Av
L(e_1)=(a,c)^T
L(e_2)= (b,d)^T

Question 126

L5 PROOF
Let V ⊂ Rⁿ be a vector subspace. Then there exists an orthonormal basis in V .

there maybe a question on the exam

Answer 126

Given an orthogonal basis→ orthonormal: by normalising vectors (GRAM SCHMIDT PROCESS exam)

If e₁,…,eₖ ∈ Rⁿ is an ORTHOGONAL BASIS in V then vectors
e₁~=e₁/ |e₁|
,…,
eₖ~=eₖ/ |eₖ|
form an orthonormal basis.
SUFFICIENT TO PROVE AN ORTHOGONAL BASIS EXISTS
Let u₁,…,uₖ be a basis of V , where k <= n is the dimension of V.
construct an orthogonal basis e₁,…,eₖ s.t

span{u₁,…,uᵢ}= span{ e₁,…,eᵢ} for all i=1,..,k

FIRSTLY
e₁=u₁
look for vector e₂ in the form e₂=u₂ + a₂,₁e₁
( e₁*e₂ =0 ,
by linearity
a₂,₁= -(u₂.e₁)/(e₁.e₁))
check span
span{u₁,u₂}= span{ e₁,e₂} ie all poss linear combos hold here

assume we have constructed the set s.t orthogonal and span cond holds:
Suppose that the non-zero orthogonal vectors e₁,…,eₗ₋₁ s.t relation holds for i=1,…,l-1 are found. Then we set
eₗ= uₗ + aₗ,₁e₁+…+ aₗ,ₗ eₗ₋₁
where coeffs are determined by conditions
eₗe₁= eₗe₂ =…=eₗeₗ₋₁
ie we find
aₗ,₁= - (uₗe₁)/ (e₁e₁)
….
aₗ,ₗ₋₁= - (uₗeₗ₋₁)/(eₗ₋₁eₗ₋₁)
Thus we find a vector eₗ s,t span set cond hold for i=1,..,l

vector eₗ non zero- o/w vectors u₁,…,uₗ would be linearly dependent. (o/w its a linear combo already in span which contradicts linearly indep)

Continue we find orthogonal collection of non zero vectors e₁,…,eᵢ satisfying spanning cond in particular

V= span{u₁,…,uₖ}= span{ e₁,…,eₖ}
conclude that e₁, . . . , eₖ is indeed a basis of V .

Question 127

We may be given a subspace in exam…

Answer 127

any subsace of R^n has an orthonormal basis (Grau schmidt- find “projection”)

Question 128

may be given subspace in exam
Given V subset in R^3
V={(x,y,z) in R^3 x+y +z=0}

Find an OB in this subspace

Answer 128

linear eq series of sols has dim 2
find any that solves that are linearly indep
e.g
u_1 =(1,-1,0)
u_2=(0,1,-1) in V
form a basis as linearly indep
orthonormal basis:
|u_1| = root 2 = |u_2|
using Gram schmidt
e_1 = (1,-1,0)
e_2 = u_2 +ae_1
e_2 = u_2 +0.5e_1 = (0.5, -0.5,-1)

a = (u_2 *e_1)/(e_1 * e_1)= -(-1)/2=1/2

checking e_1*e_2 = 0

orthonormal
e^_1 = e_1/|e_1| = (1/root(2))
e^_2 = root(2/3) (0.5,0.5,-1)

Note
e_l = u l + a{l,1}e_1 +…+a_{l,l-1}e_{l-1}

a_{l,1} = (-u_1 e_1)/(e_1e_1) ,
….,
a_{l,l-1} = -(u_l * e_{l-1})/(\e_{l-1}e_{l-1})

Question 129

notes

Answer 129

allows to construct in and dim, a vector subspace

lengths of vector in between 2 point taking a plane orthogonal to line

diagram:
plane, orthogonal (normal to plane) w

Question 130

COROLLARY I/4 for vector space
w

Answer 130

Let V subset R^n be a vector space
then for x in R^n
there exists a unique vector w in V
s.t
|x-y| > |x-w| for all y in V

y not equal to w

moreover
if e_1,..,e_n is an orthonormal basis of V then
w= sum_{i=1,n} (x* e_i) e_i

defining projection choosing a different basis doesnt change the unique vector

Question 131

PROOF
Let V subset R^n be a vector space
then for x in R^n
there exists a unique vector w in V
s.t
|x-y| > |x-w| for all y in V

y not equal to w

moreover
if e_1,..,e_n is an orthonormal basis of V then
w= sum_{i=1,n} (x* e_i) e_i

Answer 131

Show bector x-w is orthogonal to subspace V

we claim x-w orthogonal to V

(x-w)*y = 0 for all y in V

IFF
(x-w)*e_i=0 fpr all i=1,…k as all y are decomposed as

checking

by linearity
orthoormal
pythagorean thm
strictly positive
corollary 1.4 contradiction……

Question 132

e.g space of polynomials and construct an orthonormal basis

Answer 132

for example construct OB for space of polys with degree at most 2
P_2={P_n(t)=a_2t^2+a_1t+a_0 a_i in R}

we use standard inner product <f,g> integral
BASIS POLYNOMIALS
p_1(x)=1
p_2(x) =x
p_3(x) = x^2

e¬_1 = p_1(x)/ |p_1(x)| = 1/ sqrt(<1,1>) =1

e_2= x-0.5*1 =x-0.5

e¬_2 = root(2) (2x-1)
e_3=x^2-(2/3)x +(1/6) normalise by dividing by sqrt(<e_3,e_3>)

Question 133

translations x to x+a example

Answer 133

preserve length thus euclidean isometry - not linear?
T:R^n to R^n be an euclidean isometry
T(0)=0
then T is an orthogonal transform

Question 134

sketch proof that EI preserve distance

Answer 134

|T(u)-T(v)|^2 = |u-v|^2 for all u,v in R^n

(T(u)-T(v))(T(u)-T(v)) =
|u|^2 - 2 |uv| +|v|^2
properties linearity symmetry
= |T(u)|^ - 2T(u)T(v) + |T(v)|^2
cancel out if set v=0 or u=0 so
T(u)-T(v) = u*v for all u,v

NOW SHOW LINEAR MAP: take orthonormal basis
then
T(e_1),…,T(e_n) is also an orthonormal basis
………missing