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Lecture

# Lecture17-18f.pdf

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University of Toronto Scarborough

Mathematics

MATB24H3

Sophie Chrysostomou

Winter

Description

University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT B24S Fall 2011
MAT B24
Lectures 17-18.
THEOREM: Properties of Ax for an Orthogonal Matrix ALet A be an
orthogonal n × n matrix and x and y are column vectors in R . Then:
1. (Ax) ▯ (Ay) = x ▯ y Preservation of dot product.
2. ||Ax|| = ||x||Preservation of length.
3. the angle between nonzero vectors x and y equals the angle between
Ax and Ay
proof.
T T T T T T T
1. (Ax)▯(Ay) = (Ax) Ay = (x A )Ay = x (A A)y = x Iy = x y = x▯y
2. Since (Ax) ▯ (Ax) = x ▯ x then ||Ax|| = ||x|| or ||Ax|| = ||x||.
3. If the angle between the nonzero vectors Ax and Ay equals θ then
(Ax) ▯ (Ax) x ▯ x
cosθ = =
||Ax|| ▯ ||Ay|| ||x||||y||
Showing that θ is also the angle between the nonzero vectors x and y.
1 THEOREM: Orthogonality of Eigenspaces of a Real Symmetric Matrix:
Let A be a real symmetric matrix and λ and λ be1distinc2 eigenvalues of
A. Then the eigenspaces E λ1 and E λ2 are orthogonal.
proof. : To show that E λ and E λ are orthogonal we must prove v and 1
1 2
v 2re orthogonal for all v ∈ E1 λ1 and v ∈2E λ2.
Let v 1 E λ1 and v ∈2E λ2. Then
Av = λ v and Av = λ v .
1 1 1 2 2 2
Also,
λ (v ▯v ) = (λ v )▯v = (Av )▯v = (Av ) v = (v A )v = (v A)v = v (Av ) T T
1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2
= v (λ v2) 2 λ (v 2 ) = λ2(v ▯ v2) 1 2
1 1
Thus
λ (v ▯ v ) = λ (v ▯ v ) or (λ − λ )(v ▯ v ) = 0
1 1 2 2 1 2 1 2 1 2
Since λ ▯= λ then λ − λ ▯= 0 thus v ▯ v = 0
1 2 1 2 1 2
This brings us to another theorem:
THEOREM: Fundamental Theorem of Real Symmetric Matrices: Every
−1
real symmetric matrix A is diagonalizable. The diagonalization C AC = D
can be achieved by using a real orthogonal matrix C.
2
4 2 2
EXAMPLE: Let A = 2 4 2 . Find an orthogonal matrix C that diag-
2 2 4
onalizes A.
solution. det(A − λI) = (λ − 2) (λ − 8). Therefore A has only two eigen-
values1 λ = 2 a2d λ = 8.
−1 −1 1
You may verify thλt= sp 1 , 0 and Eλ = sp 1
1 1
0 1 1
We ﬁnd an orthonormal basis for each of these eigenspaces using the
Gram-Schmidt process.
√ √
−1/√2 −1/√6
So 1/ 2 ,−1/ 6 is an orthonormal basλ1 for E
√
0 2/ 6
√
1/ 3
√
and 1/√3
1/ 3
is an orthonormal basis for E
2
√ √ √
−1√ 2 −1/√6 1/ √ 2 0 0
Thus: C = 1/ 2 −1/ 6 1/ 3 and C1AC = 0 2 0
√ √
0 2/ 6 1/ 3 0 0 8
3 DEFINITION: A linear transformation T : R → R is orthogonal if it
n
satisﬁes T(v) ▯ T(w) = v ▯ w in R .
n n
THEOREM: A linear transformation T : R → R is orthogonal ⇐⇒
the standard matrix representation A of T is an orthogonal matrix.
proof. of (=⇒)
n
T is an orthogonal linear transformation of R with standard matrix repre-
n
sentation A =⇒ for all x,y ∈ R we have:
T T T T
x ▯ y = T(x) ▯ T(y) = Ax ▯ Ay = (Ax) (Ay) = x (A A)y = x ▯ (A A)y.
Thus:
T T T n
0 = x ▯ (A Ay) − (x ▯ y) = x ▯ (A Ay − y) = x ▯ (A A − I)y for all x,y ∈ R
T
If we let x = (A A − I)y we get that:
T T T
0 = x ▯ ((A A − I)y) = (A A − I)y ▯ (A A − I)y
T T
thus (A A − I)y = 0 for all y and so A A = I.
proof. (⇐=)
If the standard matrix rep of T is the orthogonal matrix A then for all
n
x,y ∈ R we have:
T T T T T
T(x) ▯ T(y) = Ax ▯ Ay = (x A )Ay = x (A A)y = x y = x ▯ y
4 EXAMPLE: Determine if T : R −→ R deﬁned by
T([x,y,z,w]) =√1 [2x − y,2y + x,2z − w,2w + z]
5
is an orthogonal linear transformation.
solution. T is a linear transformation (verify it!) with:
√1
T([1,0,0,0]) = [2,1,0,0]
5
T([0,1,0,0]) =√1 [−1,2,0,0]
5
1
T([0,0,1,0]) =√ [0,0,2,1]
5
1
T([0,0,0,1]) =√ [0,0,−1,2]
5
2 −1
√5 √5 0 0
√ √
5 5 0 0
Therefore T has a standard matrix representatio 0 = 0 √ √1
15 2
0 0 √5 √5
T
Note that A A = I therefore A is orthogonal and thus T is orthogonal.
5 The Projection Matrix
THEOREM: Projection of b on the Subspace W: Let W = sp(a ,a ,.1.a2) k
be a k-dimensional subspace of R . Let also, A be the matrix with ith
column the column vector a i If b ∈ R , then the projection of b on W is
T −1 T
bW = A(A A) A b
proof. By construction W = column space of A, thus all vectors in W
have the form Ax where x ∈ R k
We know that b = b W + b W⊥ where b W ∈ W and b W ⊥ ∈ W .
Therefore bW = Av for some vector v.
In addition, we have that bW⊥ = b − Av, thus b − Av is orthogonal
to every vector in W.
So for all vectors x ∈ R we have:
0 = Ax▯(b−Av) = (Ax) (b−Av) = x A (b−Av) = x (A b−A Av) = x▯(A b−A Av) T T
So:
0 = x ▯ (A b − A Av) for all x ∈ Rk
T T T T T −1 T
This implies that 0 = (A b−A Av) and A b = A Av or v = (A A) A b
T
[A A is invertible because A has rank k since it has k linearly independent
columns and so A A is a k × k matrix with rank k, thus invertible]
Finally, W = Av = A(A A) −1A b as wanted.
If we let P = A(A A) −1A then we call P the projection matrix for the
subspace W
6 1 Complex Numbers
In this lecture we are going to describe a number system C (the “complex
numbers”), g√neralizing the real numbers R, and large enough to give a
meaning to −1. In fact we have already mentioned them in our discussion
of ﬁelds and vector spaces.
DEFINITION: The set of complex numbers C is the set of all numbers
of the form x + iy where x,y ∈ R equipped with the following operations
1. Addition: If z1= x +1iy ∈1C and z = x 2 iy 2 C th2n we deﬁne
def
z1+ z 2 x + 1 + i2y + y 1 ∈ C2
2. Multiplication: If z = x + iy ∈ C and z = x + iy ∈ C then we
1 1 1 2 2 2
deﬁne
def
z1 2= x x1− 2 y +1 2x y + 1 2 ) ∈1C2
NOTE 1: If z = 0+i1 = i then by the above deﬁnition of multiplication, we
2
have: z = zz√= (0+i1)(0+i1) = (0−1)+i(0+0) = −1. Thus we see that
2
i = −1 or −1 = i
NOTE 2: It was proven in assignment 1 that this set with these operations
is a ﬁeld.
NOTE 3: We may deﬁne the scalar multiplication on C by: If r ∈ R and
z = x + iy ∈ C the

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