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If the product of matrices say $\ \hat{N} \hat{M} $ is symmetric $\ \left(\hat{N}\hat{M} \right)^T= \hat{N}\hat{M} $ (where T means transpose ) and $\ \hat{N} $ has the form of $\ \hat{N}=\begin{pmatrix} 0&\textrm{Identity Matrix}\\\textrm{Identity Matrix}&0\end{pmatrix} $. How can we prove that eigenvectors of the Matrix $\ \hat{M} $ ($\ \hat{M} \vec v=\lambda\vec v $) have the orthogonality property. $\ (\vec v)^T \hat{N}\vec v=\delta_{ij} $.

Thank you very much.

It is the equation (16) from this paper

https://www.dropbox.com/scl/fi/m0l566pi2z60boi0zadb4/1-s2.0-016521259290033X-main.pdf?dl=0&oref=e&r=AAWY_3uH7_BvhLaX5VHAcnQ8sGjn_rOO7mrMuASafjK9oMn9H9cI9x99X8WnZ8cwRsw_sy3sx9tlXQbYSezWSu7GmwWTvA-JWm6E2cdTZr1L3sy_aBJWYQVdPEaD5SZ8C6sj8b1duBzzvBe7NUzQjYDtz4TSpgqsBlpLgkWjtsqFnIDm_bXav9-AUBHftXBicXU&sm=1

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    Are the matrices real or complex? Or any other field? – 2017-02-16
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    Thank you very much for your respond. I came upon to this problem in some crystallography paper. And I think I forgot the linear algebra course that I had in university. – 2017-02-16
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    https://www.dropbox.com/s/nsyz59j5l1enfxf/1-s2.0-016521259290033X-main.pdf?dl=0 – 2017-02-16
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    The matrices can be considered real, I don't think that will change anything. – 2017-02-16
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    another thing is unclear. you wrote M has the form N= . which one has that form N or M? – 2017-02-16
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    Sorry, It should've been N there. I fixed it. – 2017-02-16
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    OK there are still problems with your question. the definition of the orthogonality property doesn't make sense. one side of the equality has indices while the other side doesn't – 2017-02-16

1 Answers 1

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Let $v_{i},v_{j}$ be eigenvectors of $M$ with eigenvalues $\lambda_{i},\ \lambda_{j}$ respectively. We assume $\lambda_{i}\neq \lambda_{j}$ . We note $u^{T}Nv$ by $_{N}$. Notice that $N^{T}=N$ so we have $NM=(NM)^{T}=M^{T}N^{T}=M^{T}N$. We observe that $$\begin{eqnarray*} \lambda_{i}\left\langle v_{i},v_{j}\right\rangle _{N} & = & v_{i}^{T}M^{T}Nv_{j}\\ & = & v_{i}^{T}NMv_{j}\\ & = & v_{i}^{T}N\lambda_{j}v_{j}\\ & = & \lambda_{j}\left\langle v_{i},v_{j}\right\rangle _{N} \end{eqnarray*}$$ This implies $\left\langle v_{i},v_{j}\right\rangle _{N}=0$. If $v_{i}$ is an eigenvector then so is $cv_{i}$ therefore If we want $\left\langle v_{i},v_{i}\right\rangle _{N}=1$ we need to choose normlized vectors to begin with.