Analysis of Orthogonal Matrices, Eigenvalues, and Quadratic Forms

This assignment delves into key concepts of linear algebra, providing detailed solutions to problems involving orthogonal matrices, eigenvalues, and quadratic forms. Part 1 focuses on the properties of orthogonal matrices, including their relationship to identity matrices and eigenvalues. It als...Read More

Part 1
Question one
Orthogonal matrix
For a matrix to be orthogonal; Transpose¿ inverse
The conditions are
a∧bshould each be equal to 0.5 so that a+ b=1=b+ a and a−b=b−a=0 this makes the
matrix an identity which is orthogonal.
If it is an identity matrix it will as well be a diagonal matrix which is also an orthogonal
matrix.
The matrix will only be orthogonal if a and b have real values.
Question two
If A is symmetric orthogonal then ± 1 are the only possible eigen values
Consider | Av|=¿ v∨¿ for any v ,suppose a is orthogonal and v ≠ 0is an eigenvector having
eigen value λ
av =λv implying |v|=|av|=| λ|∨v∨¿ hence |λ|=1
Which gives the value of λ as either 1∨−1
Question three
The quadratic form 2 x2 +5 y2 +5 z2 + 4 xy−4 xz−8 yz has an associated matrix that can be
orthogonally diagonalized
[
2 2 −2
2 5 −4
−2 −4 5
¿
−λ3+12 λ2−21 λ+10
λ1=1 hence v1=(−2 1 0)
λ2=1 hence v2=(2 0 1)
λ3=10 hence v3=(−1−2 2)
being that λ1=λ2 , v1 is perpendicular ¿ v3∧v2 ¿ v3∧v1 ¿ v2 this can be solved in a linear
combination of
e1= 1
√ 5 ( 2
0
1) ∧e3= 1
3∗(
−1
−2
2
)

So e2=e3∗e1=[
−2
3 √5
√5
3
4
3 √5
]
Then [ x'
y'
z' ]=
( 2
√5
−2
3 √5
−1
3
0 √5
3
−2
3
1
√5
4
3 √5
3
3
)−1
( x
y
z )=
( 2
√5 0 1
√5
−2
3 √5
√5
3
4
3 √5
−1
3
−2
3
2
3
) ( x
y
z )
( 2
√5
−2
3 √ 5
−1
3
0 √5
3
−2
3
1
√5
4
3 √ 5
2
3
) ( 2 2 −2
2 5 −4
−2 −4 5 ) ( 2
√5 0 1
√5
−2
3 √5
√ 5
3
4
3 √5
−1
3
−2
3
2
3
)=
(1 0 0
0 1 0
0 0 10 )
Now expressing q in terms of the new variables gives
Q=xr 2 + yr 2+10 zr 2
Question four
To show that a quadratic equation represent ellipsoid
4
3 x2+ 4
3 y2 + 4
3 z2+ 4
3 xy + 4
3 xz + 4
3 yz=1
The general form of an ellipsoid is
A x2+ B x2+c z2 + Dyz+ Ezx + Fxy+ Hy +Jz + k=0
The matrix
M =
[ A F
2
D
2
F
2 B D
2
E
2
D
2 C ] should be positive definite i.e.

A>0 , AB > F2
4 ∧det ( M ) =0
Now proving our equation
M =
[ 4
3
4
6
4
6
4
6
4
3
4
6
4
6
4
6
4
3 ]Now A= 4
3 >0
AB=( 4
3 )
2
=16
9
While F2
4 = 4
9 = 4
9
Hence AB> F2
4
The det ( M )= 4
3∗
| 4
6
4
6
4
6
4
3 |− 4
6 |4
6
4
6
4
6
4
3 |+ 4
6 ∗
|4
6
4
3
4
6
4
6 |
This gives
16
27 − 8
27 − 8
27 =0
position and length of principal axis
getting matrix
M =
[ 4
3
4
3
4
3
4
3
4
3
4
3
4
3
4
3
4
3
]Now we obtain the eigen values

[ 4
3 − λ 4
3
4
3
4
3
4
3 −λ 4
3
4
3
4
3
4
3 −λ ]
( 4
3 −λ ) [ 4
3 − λ 4
3
4
3
4
3 −λ ] − 4
3 [ 4
3
4
3
4
3
4
3 − λ ] + 4
3 [ 4
3
4
3 − λ
4
3
4
3 ]
Which gives the equation
λ3+ 4
3 λ2−4 λ
From the equation, we get the eigen values as
λ1=4 , λ2=0 ,∧λ3=0
And the corresponding eigen vectors as
v1 =(1 , 1, 1)
v2=(−1 , 0 , 1)
v3 =(−1 , 1 ,0)
We then normalize the eigen vectors to obtain a matrix
[ 1
√ 3
−1
√2
−1
√2
1
√ 3 0 1
√2
1
√ 3
1
√2 0 ]
We get D= [4 0 0
0 0 0
0 0 0 ]
And the equations
y1= 1
√3 x + 1
√3 y+ 1
√3 z
y2=−1
√ 2 x + 1
√ 2 z
y3=−1
√ 2 x + 1
√2 y

Now the principal axis have the equation
y1= 1
√3 x + 1
√3 y+ 1
√3 z
And length
|| y1||= √((¿ 1
√3 )
2
+( 1
√3 )
2
+( 1
√3 )
2
) ¿
Then || y1||=1
Part 2
Question 1
Inner product is defined in the form of suppose u and v are vectors then
u . v=u1 v1+u2 v2 +… . un vn
In B ( u , v ) = ( u1 , u2 ,u3 ) [ 1 1 1
1 2 0
1 0 3 ] ( v1
v2
v3 )
In the multiplication of a matrix
We have u1∗1+u2∗1+u3∗1, u1∗1+u2∗2+u3 +0 , u1∗1+u2∗0+u3∗3
This is a form of an inner product as it takes on the form given above and gives the eventual
answer as a scalar.
Length of a vector
a=(1 , 1 , 0)
||a||= √a . a= √12 +12+ 02= √2
For b=(0 ,1 , 1)
||b||= √b .b= √2
Finding the angle between the vectors
cosθ= a .b
||a||∗||b|| , where a . b=inner product
Hence obtaining the angle with respect to the inner product gives
a . b=¿cosθ∗¿|a|∨¿|b|∨¿ where θ is theangle between the vectors
But a . b=0+1+0=1

Hence cosθ= 1
√2∗√ 2 =1
2 which means θ=600
Question 2
Using the QR decomposition where R=B=upper triangular ¿ Q= A=orthogonal ¿
AB= [1 1 1
1 3 9
1 9 81 ]
a1= ( 1,1,1 )T , a2= ( 1,3,9 )T ,a3= ( 1,9,81 )T
The vectors of a are column vectors
Now conducting the Gram-Schmidt procedure we have
u1=a1=(1,1,1)
e1= u1
||u1||= 1
√3∗( 1,1,1 )=( 1
√3 , 1
√3 , 1
√3 )
u2=a2− ( a2 . e1 ) e1
¿ ( 1,3,9 )− 13
√3 ( 1
√3 , 1
√3 , 1
√3 )=(−10
3 ,− 4
3 , 11
3 )
e2= u2
||u2||= 3
√237∗(−10
3 ,− 4
3 , 11
3 )=( −10
√237 , −4
√237 , 11
√237 )
u3=a3− ( a3 . e1 ) e1− ( a3 . e2 ) e2
( 1 , 9 ,81 )− 91
√3 ( 1
√ 3 , 1
√3 , 1
√3 )− 845
√237 ( −10
√237 , −4
√237 , 11
√237 )=( 1498
237 ,− 1676
237 , 2713
237 )
e3= u3
||u3||= 237
2713 u3
¿ 237
2713∗( 1498
237 ,− 1676
237 , 2713
237 )=( √ 3
10 ,− √ 39
100 , 1)
Q= [ e1 , e2 , e3 ]=
[ 1
√3
−10
√237 √ 3
10
1
√3
−4
√237 √ 39
100
1
√3
11
√237 1 ]

R=
[a1 . e1 a2 . e1 a3 . e1
0 a2 . e2 a3 . e2
0 0 a3 . e3 ]=
[ 3
√3
13
√3
91
√3
0 77
√237
845
√237
0 0 523
6
]
Question 3
x1= [1
1
1 ], x2=
[1
3
9 ], x3= [ 1
9
81 ]
v1 =x1
v2=x2 − x2
T∗v1
v1
1 v1
v1=¿[
1
3
9
] - (1 3 9)(
1
1
1
)
¿ ¿
v3 =x3− x3
T v1
v1
T v1
v1− x3
T v2
v2
T v2
v2
[ 1
9
81 ]−
( 1 9 81 ) [ 1
1
1 ]
[ 1 1 1 ] [ 1
1
1 ] ∗
[ 1
1
1 ]−
[ 1 9 81 ]
[ −10
3
−4
3
14
3
]
[ −10
3
−4
3
14
3 ] [ −10
3
−4
3
14
3
] ∗
[ −10
3
−4
3
14
3
]
v3 =
[ 2884 /39
761/39
−3736 /39 ]

Orthogonal basis W =span {
[1
1
1 ] [ −10
3
−4
3
14
3 ][ 2884
39
761
39
−3736
39 ]}
othornormal basis R=span { v1
||v1|| , v2
||v2|| , v3
||v3|| }
¿∨v1∨¿= √12+ 12+12 = √3
||v2||= √ (−10
3 )2
+¿ (−4
3 )2
+¿ (14
3 )2
= √ 312
9
¿∨v3∨¿=4720/39
R=span
{[ 1
√3
1
√3
1
√3
] [ − 10
3
√ 312
9
− 4
3
√ 312
9
14
3
√ 312
9
] [ 721
1180
761
4720
3736
4720
] }
¿
[ 1
√3
1
√3
1
√3
] [ −10
√312
−4
√312
14
√312
[ 721
1180
761
4720
3736
4720
] ]
Question 4
a) u=(4 , 2 ,1)
v1 =( 1
3 , 2
3 ,−2
3 )

v2=( 2
3 , 1
3 , 2
3 )
The orthogonal projection of u on to the plane spanned by vectors v1 ∧v2
Remark
v1 . v1= ( 1
3 )2
+( 2
3 )2
+ (−2.3 )2=1
v1 . v2= ( 1
3∗2
3 )+ ( 2
3 ∗1
3 )+( −2
3 ∗2
3 )=0
v2 . v2= ( 2
3 )2
+( 1
3 )2
+ ( 2
3 )2
=1
In this case we get that v1 ∧v2 areperpendicular with length √1=1
Therefore, the vectors
u1=1∗v1= ( 1
3 , 2
3 ,−2
3 )∧u2=1∗v2 =( 2
3 , 1
3 , 2
3 ) form the orthogonal basis of the plane.
therefore
u . u1= 4
3 + 4
3 − 2
3 =2
And
u . u2= 8
3 + 2
3 + 2
3 =4
In conclusion, the projection of u can be obtained by the formula
u¿∨¿= ( u . u1 ) u1+¿ ¿(u . u2 ¿ u2
¿ 2∗( 1
3 , 2
3 ,−2
3 )+ 4∗( 2
3 , 1
3 , 2
3 )=( 10
3 , 8
3 , 4
3 )
b) obtain the cross product
v3 =v1∗v2=
[ i j k
1
3
2
3
−2
3
2
3
1
3
2
3 ]= 2
3 i− 2
3 j−1
3 k
Now computing the component orthogonal to the plane is the projection of u onto v3
parallel to the plane
pr ⊥v3
u= ( u .u3 ) u3=1∗u3=( 2
3 ,− 2
3 ,−1
3 )