Question

Prove that eigen values of hermitian matrices are real.

Answer 1

Step-by-step explanation:

Let A be a given Hermitian matrix, then Aθ=A

Let λ be the characteristic root of the matrix A with corresponding characteristic vector X then AX=λX (1)

Taking transpose conjugate of both the sides of (1) we get

∴(AX)θ=(λX)θ∴XθAθ=λ¯XθXθA=λ¯Xθ

Post - multiplying by X, we get

XθAX=λ¯XθXXθλX=λ¯XθXλXθX=λ¯XθX(λ−λ¯)XθX=0

Since X is the non-zero vector, XθX≠0 →λ=λ¯ which shows that λ is real.

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