Disentangling orthogonal matrices

MATH1
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MATH2
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Problem 1.1
1.1. ¿ ⟨ x , y ⟩∨¿∨⟨ λy , y ⟩∨¿∨λ∨⟨ y , y ⟩SS
¿∨λ∨∥ y ∥ 2=∥ λy ∥∥ y ∥=∥ x ∥∥ y ∥
In a linearly independent case: If x−λy≠0 and y−λx≠0 for all λ∈K, then also x,y≠0, and
¿ 0< ⟨ x+ λy , x + λy ⟩
¿ ⟨ x , x+ λy ⟩ + λ ⟨ y , x + λy ⟩
¿ ⟨ x , x ⟩ + ⟨ x , λy ⟩ + λ ⟨ y , x ⟩ + λ ⟨ y , λy ⟩
¿ ∥ x ∥2+ λ ⟨ x , y ⟩ +λ ⟨ x , y ⟩ + λλ ∥ y ∥ 2
¿ ∥ x ∥2+ 2 R ( λ ⟨ x , y ⟩ ) +|λ|
2 ∥ y ∥2
If ⟨x,y⟩=0 the Cauchy–Schwarz inequality is trivial, so we assume that ⟨x,y⟩≠0.
Then let λ:=tu where u:=⟨x,y⟩|⟨x,y⟩|,
so that : λ¯⟨x,y⟩ ¿ t ⟨ x , y ⟩
⟨ x , y ⟩|⟨ x , y ⟩|= t|⟨x,y⟩| and |λ|2=t2.
Hence,
0<∥ x ∥2 +2 t |⟨ x , y ⟩|+t2 ∥ y ∥2
¿ (∥ y ∥t+|⟨ x , y ⟩|∥ y ∥ )2
+∥ x ∥2− (|⟨ x , y ⟩|∥ y ∥ )2
By choosingt=−|⟨ x , y ⟩|
∥ y ∥2 , we obtain that
¿ |⟨ x , y ⟩|2
∥ y ∥2 =∥ x ∥2
Hence proven. Since ¿ x , x >¿ ∥ x ∥2 (Mackey, Mackey and Tisseur 2005).
1.2. Problem
If ⟨ x , y ⟩=0in an inner-product space, then
∥ x + y ∥2=∥ x ∥2+∥ y ∥2
which in, R2 we recognize as
a2+b2=c2,
with a,b,c
 the sides of a right-angled triangle.
 If we define ⟨ x , y ⟩ := 1
4 ( ∥x + y ∥2−∥ x− y ∥2)

MATH3
in R2using the polarization identity , we see that
¿ x , y ≥ 1
4 ( ( x1+ y2 )2+ ( x2 + y2 )2
)− 1
4 ( ( x1− y1 ) 2+ ( x2− y2 )2)
= 1
4 ( x1
2+2x1 y2+ y1
2+ x2
2+2x2 y2+ y2
2)− 1
4 ( x1
2−2x1 y1+ y1
2+ x2
2−2 x2 y2+ y2
2)
¿ xy + xy
Which is the standard dot product.
1.1. Sn-1 can be parameterized by ⃗ x ∈ sn−1 ↔ ( cosθn−1 , sinθn−1⃗ V n−1 )
Where;⃗ V n−1 ∈ sn−1∧θn−1 ∈ [ 0 , π ] ( 0,2 π ) for s1
¿ sin t , cos t >¿ ∫
− π
π
sin t cos t dt
( 0<t <2 π ) =∫
0
2 π
sin t cos t dt
¿ 2 π
1.2. We can prove that the parameterized sn−1can be parameterized by:
ds2n-1=d θ2 +Si n2 θn−1 d2
the functions sin t and cos t are orthogonal as
¿ sin t , cos t >¿ ∫
− π
π
sin t cos t dt
¿ 1
2 sin2 t∨¿−π
¿ 0−0
¿ 0
1.3. We know that
dS20 = 0
to prove that ds1
2=d φ2
and
that ds2
2=d θ2 +sin2 ❑θd φ2
the known results for S1(a unit circle) and S2 (unit sphere in R2)
is done by the same intergration;
ds1
2=∫
0
φ2
sin t cos t dt
¿( ( 0,1 ) , ( 0,1 ) )
which is a coordinate for the a circle.

MATH4
While
ds2
2= ∫
θ2
sin2 φ2
sin t cos t dt
¿ ( 0,1,1 )
Which is a unit sphere.
Problem 2.1
Proof
V is an n-dimensional vector space, thus its basis B = {V1,…,Vn}
Where every Vi is a vector space in V
By defining a map T: V →R n we send every vector v∈V to the coordinate vector [v] B based
on the basis
That is v=c1v1+⋯+cnvn with c1,…,cn∈R,
The coordinate vector with respect to the basis becomes
[v]B =
C 1
C 2
Cn
∈ R n
Mapping T:V → R n is defined by
T (v) =
C 1
C 2
Cn
from the property of coordinate vectors we conclude that map T is a linear transformation.
Thus, to prove that isomorphism we show that T is bijective
If v ∈N (T), then 0=T(v)=[v]B.
Coordinate vector of v= 0 hence
v=0v1+⋯+0vn=0.
Therefore N (T) = {0} N (T) = {0}, hence T is injective
Showing that T is surjective
Let a =
C 1
C 2
Cn

MATH5
Be an a arbitrary vector in the vector space
Thus consider v: = a1v1+⋯+anvn in V
The linear transformation T by definition becomes
T(v) = [V] B =
C 1
C 2
Cn
= a
Hence T is surjective
Therefore, T: V → R n is a bijective linear transformation thus making T to be an isomorphism.
In conclusion Rn and V are isomorphic (Zhang and Singer 2017).
Problem 2.2
(A, B) = Tr (A T . B)
If V is a n-dimensional vector space and B is the matrix of the inner product that is relative to
basis B
Then u, v ∈ V
(u, v) xTAy
x and y a coordinate vectors
< x, y> = xT y = ∑ xi . yi , x, y ∈ R n .
Standard linear product becomes
Thus
(Chandra, Lal, Raghavendra, and Santhanam 2013)
Problem 2.3
Proof:
Space of orthogonal matrices is a subset of √n Sn
QTQ = QQT = I
QT = Q-1
A n × n orthogonal matrices form an orthogonal group O (n)

MATH6
Considering an (n + 1) × (n + 1) orthogonal matrices with matrix given as
SO (n) is a subgroup of SO (n + 1) hence
The orthogonal transformation ∈ √n Sn
Problem 2.4
Proof: antisymmetric matrices is an Abelian group
(An, +) ≃ ( R n (n−1)/2 , +)
A(0) = I, A (t)=I for all t =0
We consider A/ d t (0) + d AT / dt (0) =0
(An, +) = ( R n R (n−1)/2 , +),
(An, +) = R n (n−1)/2 , +)
However, the matrix exponential map of the antisymmetric matrix is an orthogonal matrix
Hence, (Sn, +) ≃ (R n (n−1)/2 , +)
<A, S> = 0 ∀A ∈ An, S ∈ Sn
PROBLEM 2.5
The space M (G )of matrix is defined by M. G. MT =G,
The transpose of the M matrix m × n matrix A = [aij ]
n × m matrix B = [bij ] hence satisfying G.G =G2= ∥n×n
The manifold dimension becomes dim(M) = n(n − 1)/2,
And, M ∈ M ∩ SL(n) ⇒ M = exp (A · G), A ∈ An
Hence showing that the space of matrix is a matrix satisfying G · G = G2

MATH7
References
Chandra, P., Lal, A.K., Raghavendra, V. and Santhanam, G. (2013). Notes on Mathematics-
1021. Funded by Mhrd-India, pp.9-46.
Mackey, D.S., Mackey, N. and Tisseur, F. (2005). Structured factorizations in scalar product
spaces. SIAM journal on matrix analysis and applications, 27(3), pp.821-850.
Zhang, T. and Singer, A. (2017). Disentangling orthogonal matrices. Linear algebra and its
applications, 524, pp.159-181.