In a Hermitian Matrix, the Eigenvectors of Different Eigenvalues are Orthogonal

This is an elementary (yet important) fact in matrix analysis.

Statement

Let M be an n\times n complex Hermitian matrix which means M=M^* where * denotes the conjugate transpose operation. Let \lambda_1, \neq \lambda_2 be two different  eigenvalues of M. Let x, y be the two eigenvectors of M corresponding to the two eigenvalues \lambda_1 and \lambda_2, respectively.

Then the following is true:

\boxed{\lambda_1 \neq \lambda_2 \iff \langle x, y \rangle = 0.}

Here \langle a,b\rangle denotes the usual inner product of two vectors x,y, i.e.,

\langle x,y \rangle := y^*x.

Proof

It is given that

Mx=\lambda_1x,
My=\lambda_2y.

Since M=M^*, it follows that

 \langle x,My \rangle =y^*M^*x=y^*Mx=y^*\lambda_1x=\lambda_1y^*x=\lambda_1 \langle x,y \rangle .

However, we have

\langle x,My \rangle = \langle x, \lambda_2y \rangle =\lambda_2  \langle x,y \rangle .

Therefore,

\lambda_1 \langle x,y \rangle =\lambda_2 \langle x,y \rangle 
\implies (\lambda_1-\lambda_2) \langle x,y \rangle =0.

Therefore, \lambda_1 \neq \lambda_2 \implies \langle x,y \rangle =0,

and

\langle x,y \rangle \neq 0 \implies \lambda_1=\lambda_2.

Thus the eigenvectors corresponding to different eigenvalues of a Hermitian matrix are orthogonal. Additionally, the eigenvalues corresponding to a pair of non-orthogonal eigenvectors are equal.

\square

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