Hermitian Matrices have Real Eigenvalues
• a matrix is Hermitian if it is equal to its own conjugate transpose
• conjugate transpose of a matrix is its transpose with each element complex conjugated
• denoted by a superscript dagger, 𝐴 = 𝐴! (superscript 𝐻 and ∗ are also common)
• Examples of (non) Hermitian matrices:
• 𝐴=
1
0
2
1
is not Hermitian because 𝐴! = (𝐴)̅ " =
1
2
• 𝐵=
3
1 + 5𝑖
0
≠𝐴
1
1 − 5𝑖
3
2 "=
is Hermitian because 𝐵! = (𝐵)
1
1 + 5𝑖
• all real symmetric matrices are Hermitian
1 − 5𝑖
=𝐵
1
Proof
1. Let 𝜆 be an eigenvalue of a Hermitian matrix, 𝐴, that corresponds to eigenvector, 𝒙. Then
𝐴𝒙 = 𝜆𝒙
(1)
2. Now, left multiply both sides of Eq. 1 by the conjugate transpose of 𝒙
𝒙! 𝐴𝒙 = 𝒙! 𝜆𝒙
(2)
= 𝜆(𝒙! 𝒙)
(3)
= 𝜆((
𝒙" 𝒙)
(4)
=𝜆 𝒙
(5)
3. Take the conjugate transpose of the LHS of Eq. 2
(𝒙! 𝐴𝒙)! = 𝒙! 𝐴! (𝒙! )!
recall 𝐴 is Hermitian, 𝐴! = 𝐴
(6)
= 𝒙! 𝐴! 𝒙
(7)
= 𝒙! 𝐴𝒙
(8)
= 𝜆𝒙! 𝒙
(9)
=𝜆 𝒙
(10)
4. We know that the conjugate transpose of the LHS of Eq. 2 should be equal to the conjugate transpose of Eq. 5
𝜆 𝒙 = (𝜆 𝒙 )!
= 𝜆̅ 𝒙
∴ 𝜆 = 𝜆̅
(11)
(12)
(13)