Linear Algebra Homework: Matrices Properties and Examples

This assignment solution delves into various aspects of matrix theory, including Hermitian, skew-Hermitian, unitary, and orthogonal matrices. It provides detailed proofs and examples for properties such as the nature of diagonal elements in Hermitian and skew-Hermitian matrices. The solution explores the determination of unitary and orthogonal matrices, including demonstrating the absolute value of a unitary matrix's determinant. The document further includes the calculation of eigenvalues and eigenvectors for given matrices, offering a comprehensive understanding of matrix operations and their applications. The document covers various aspects of linear algebra related to matrices.

MATRICES
Hermitian, skew-hermitian, unitary and orthogonal matrices
Student id
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Question 1
A=
[ 3+ 4 i −5 i
−7 6−2i ]
Here ,
a=x+ iy
a=x−iy
a∧a are complex cnjugates .
A=[3−4 i 5 i
−7 6 +2i ]
Question 2
(a) We need to prove that if A is Hermitian, then the entries on the main diagonal would
satisfy
a jj=a jj
Let
a jj=x +iy
Further, assume that i , j are the main diagonal element and hence,
aij=a ji … … … … … … .. ( 1 )
Hence,
i= j
From equation1
a jj=a jj
Proved, Moreover, the conclusion can be made based on the below highlighted understanding.
a jj=x +iy
a jj=x−iy
Thus,
1

a jj=a jj
x +iy=x −iy
y=0
Therefore, the conclusion can be made that hermitian matrix A would satisfy the condition
a jj=a jj only when the imaginary element is zero or only real elements present in the main
diagonal of matrix.
(b) We need to prove that if A is Skew- hermitian, then a jj=−a jj
Let
a jj=x +iy
Further, assume that j , k are the main diagonal element and hence,
akj=−ajk … … … … … … .. ( 1 )
Hence,
a jj=−a jj
Proved, Moreover, the conclusion can be made based on the below highlighted understanding.
a jj=x +iy
a jj=x−iy
Thus,
a jj=−a jj
x−iy=−( x+iy)
x−iy=−x−iy
x=0
Therefore, the conclusion can be made that skew-hermitian matrix A would satisfy the condition
a jj=−a jj only when the real element is zero in the main diagonal of matrix.
Question 3
2

(a) Let A = [ 4 1−3 i
1+ 3i 7 ]
a=x+ iy
a=x−iy
a∧a are complex cnjugates .
A=
[ 4 1+3 i
1−3 i 7 ]
AT = [ 4 1−3i
1+3 i 7 ]
It is apparent that AT = A
Hence , the¿ be termed as Hermitian¿
(b) Let B = [ 3 i 2+i
−2+i −i ]
a=x+ iy
a=x−iy
a∧a are complex cnjugates .
B= [ −3i 2−i
−2−i i ]
BT =
[ −3 i −2−i
2−i i ]
It is apparent that BT =−B
[−3 i −2−i
2−i i ]=−
[ 3 i 2+i
−2+i −i ]=
[−3 i −2−i
2−i i ]
Hence , the¿ be termed as Skew−Hermitian ¿
3

(c) Let C =
[ i
2
√3
2
√ 3
2
i
2 ]
a=x+ iy
a=x−iy
a∧a are complex cnjugates .
C=
[ −i
2
√ 3
2
√ 3
2
−i
2 ]
CT=
[ −i
2
√ 3
2
√ 3
2
−i
2 ]
The matrix is neither hermitian nor skew-hermitian and hence, the next step is to fine whether
this is a unitary matrix.
Now,
C−1=
[ i
2
√3
2
√3
2
i
2 ]
¿ 1
det
[ i
2
√ 3
2
√ 3
2
i
2 ] [ i
2
− √ 3
2
− √ 3
2
i
2 ]
¿ 1
−1 [ i
2
− √ 3
2
− √ 3
2
i
2 ] 4

C−1=
[ −i
2
√3
2
√3
2
−i
2 ]
CT=C−1
[ −i
2
√3
2
√ 3
2
−i
2 ]=
[−i
2
√ 3
2
√3
2
−i
2 ]
It can be seen that CT=C−1 and hence, it can be said that the matrix is unitary matrix.
Question 4
 Hermitian matrix is real indicated that the matrix does not comprise any imaginary parts.
 For hermitian matrix AT = A
 A=A Because the matrix contains only real values.
 Hence, AT = A=A
For example
Let the matrix A =[ 2 4
4 3 ] is a Hermitian real matrix.
A=[ 2 4
4 3 ]
AT = [2 4
4 3 ]
Hence, AT = A=A
[2 4
4 3 ]=[ 2 4
4 3 ] =[2 4
4 3 ]
Hence, AT = A=Afor Hermitian real matrix
Question 5
5

Orthogonal matrix is a real matrix when the inverse equals its transpose.
Let assume that A is the matrix (n*n) with real entries.
Further,
AT A=I … … …… (1)
Now,
( A¿¿ T A) A−1 =I A−1 ¿
From equation 1
( A¿¿ T A) A−1 =AT ( A A−1 ) ¿
Hence, in order to satisfied the above relation, AT = A−1
Proved
Question 6
“Determine of a unitary matrix A has absolute value 1.”
Determine of a unitary (n*n) would has absolute value
1. det ( A¿ ) =det ¿
2. det ( AB ) =¿ det ( A ) det ( B ) ¿
3. det ( I ) =1
Now,
Let A is a unitary matrix and hence, A A¿=I
det ( A A¿ ) =det ( I )
det ( A ) .det ( A¿ ) =1
A A¿=1
A2=1
¿ det ( A)∨¿ 1
Proved
6

Question 7
(a) X =[ 0 1
−1 0 ]
Now, the value of transpose needs to be found.
XT = [ 0 1
−1 0 ] T
= [ 0 −1
1 0 ]
It is apparent that XT =− X
[ 0 −1
1 0 ]=− [ 0 1
−1 0 ] = [ 0 −1
1 0 ]
Therefore, the matrix X is Skew symmetric.
Determinant of matrix X
det ( X )=¿|0 −1
1 0 |=0− (−1 )=1 ¿
(b) Y = [ cosθ −sinθ
sinθ cosθ ]
Now, the value of transpose needs to be found.
Y T = [cosθ −sinθ
sinθ cosθ ]T
= [ cosθ sinθ
−sinθ cosθ ]
It is apparent that Y T =−Y
Therefore, the matrix X is Skew symmetric.
Determinant of matrix X
7

det ( Y )=¿ [cosθ −sinθ
sinθ cosθ ]=cos2 θ− (−sin2 θ )=cos2 θ+sin2 θ=1 ¿
(c) Z=
[ cosθ −sinθ 0
sinθ cosθ 0
0 0 1 ]
Now, the value of transpose needs to be found.
ZT = [cosθ −sinθ 0
sinθ cosθ 0
0 0 1 ]T
= [ cosθ sinθ 0
−sinθ cosθ 0
0 0 1 ]
The matrix would be termed as orthogonal matrix when A . AT =I
Hence, let’s find the multiplication of Z . ZT
¿ [cosθ −sinθ 0
sinθ cosθ 0
0 0 1 ]. [ cosθ sinθ 0
−sinθ cosθ 0
0 0 1 ]
¿ [ cos2 θ+sin2 θ cosθ sinθ−sinθ+ cosθ 0
sinθcosθ−sincosθ+0 sin2 θ+ cos2 θ 0
0 0 1 ]
¿ [cos2 θ+ sin2 θ 0 0
0 sin2 θ+cos2 θ 0
0 0 1 ]
cos2 θ+ sin2 θ=1
¿ [ 1 0 0
0 1 0
0 0 1 ]
It is apparent that Z . ZT =1. Therefore ,the given¿ be termed as orthogonal ¿
Question 8
Eigen values of Matrix A =?
8

(a) The Matrix A is shown below:
A=
[−2 2 −3
2 1 −6
−1 −2 0 ]
Let
A is a square matrix such that, Ax=λ x
When x ≠ 0 , then
λ=Number ¿
Step 1
 ( A−λ I)
¿ [−2 2 −3
2 1 −6
−1 −2 0 ]− λ [1 0 0
0 1 0
0 0 1 ]
¿ [−2 2 −3
2 1 −6
−1 −2 0 ]− [ λ 0 0
0 λ 0
0 0 λ ]
¿ [
−2−λ 2 −3
2 1−λ −6
−1 −2 −λ ]
Step 2
 | A−λ I ∨¿
The value of determinant of obtained matrix needs to be found.
| A−λ I∨¿
|−2− λ 2 −3
2 1−λ −6
−1 −2 −λ|
¿ (−2− λ ) {−λ ( 1− λ )− (6∗2 ) }−2 {2 (− λ )− (−1∗−6 ) }−3 {−4− (−1∗( 1−λ ) ) }
¿ (−2− λ ) {−λ+ λ2−12 }−2 {−2 λ−6 }−3 {−3− λ }
¿− ( λ−5 ) ¿
9

Step 3
 | A−λ I∨¿ 0
− ( λ−5 ) ¿
λ=5 ,
λ=−3
Therefore, the eigenvalues for the matrix A comes out to be 5,-3.
(b) The Matrix A is shown below:
A=
[ 0 1
−1 0 ]
Let
A is a square matrix such that, Ax=λ x
When x ≠ 0 , then
λ=Number ¿
Step 1
 ( A−λ I)
¿ [ 0 1
−1 0 ] −λ [ 1 0
0 1 ]
¿ [ 0 1
−1 0 ]−[ λ 0
0 λ ]
¿ [− λ 1
−1 − λ ]
Step 2
 | A−λ I∨¿
10

The value of determinant of obtained matrix needs to be found.
| A−λ I ∨¿ [ −λ 1
−1 −λ ]
¿ λ2− (−1 )
¿ λ2+1
Step 3
 | A−λ I∨¿ 0
λ2+1=0
λ= √ −1
λ=i ,−i
Therefore, the eigenvalues for the matrix A comes out to be i ,−i.
Question 9
Eigen vector of Matrix A =?
(a) The Matrix A is shown below:
A=
[ 0 1
−1 0 ]
The eigenvalues for the matrix A comes out to be i ,−i.
 Eigenvector corresponding to eigenvalue i
Step 1
( A−λ I ¿=
[ 0 1
−1 0 ]−i [1 0
0 1 ]
¿ [ 0 1
−1 0 ]−
[ i 0
0 i ]
¿ [ −i 1
−1 −i ]
11