Falak Kinney

2021-01-05

Prove that the eigenvalues of a Hermitian matrix are real.

aprovard

Skilled2021-01-06Added 94 answers

Let λ be an arbitrary eigenvalue of a Hermitian matrix A and let x be an eigenvector corresponding to the eigenvalue λ. Then we have Ax=λx.(*) Multiplying by x¯T from the left, we obtain x¯T(Ax)=x¯T(λx)=λx¯Tx=λ||x||. We also have x¯T(Ax)=(Ax)Tx¯=xTATx¯. The first equality follows because the dot product u⋅v of vectors u,v is commutative. That is, we have u⋅v=uTv=vTu=v⋅u. We applied this fact with u=x¯ and v=Ax. Thus we obtain xTATx¯=λ||x||. Taking the complex conjugate of this equality, we have x¯TA¯Tx=λ¯||x||.(**) (Note that x¯¯=x. Also ||x||¯¯¯¯¯¯¯¯¯=||x|| because ||x|| is a real number.) Since the matrix A is Hermitian, we have A¯T=A. This yields that λ¯||x||=(∗∗)x¯TAx=(∗)x¯Tλx=λ||x||. Recall that x is an eigenvector, hence x is not the zero vector and the length ||x||≠0. Therefore, we divide by the length ||x|| and get λ=λ¯. It follows from this that the eigenvalue λ is a real number. Since λ is an arbitrary eigenvalue of A, we conclude that all the eigenvalues of the Hermitian matrix A are real numbers.

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