I have the eigenvectors of a symmetric, normal, complex $n*n$ matrix. With degenerate eigenvalues and eigenvectors $\lvert m_i\rangle$. I know the Gram-Schmidt or the Householder transformation method can be used to make them orthogonal and satisfy $\langle m_i\lvert m_j \rangle=\delta_{ij}$. Which I already have tried.

But what I need is eigenvectors satisfying the following conditions:

$$\langle \overline m_i\lvert m_j \rangle=0 \quad\quad\quad if\quad i\neq j$$

$$\sum_{i=1}^{n} \frac{\lvert m_i \rangle \langle \overline m_i \lvert}{\langle \overline m_i\lvert m_i \rangle} = I_n$$

Where $\langle \overline m_i\lvert $ is the original eigenvector $\lvert m_i \rangle$ just transposed (not conjugated), and $I_n$ is the identity matrix of size $n$.

Is such thing possible and if yes, How?

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    $\begingroup$ I assume those are supposed to be bra-kets and ket-bras? But you must be missing indices $i$ and $j$ in several places of your question. Also if $m$ is the integer in the loop, the inner and outer products don't make sense since the summand will evaluate to scalars. You don't want to take inner and outer products of integers, but vectors, I presume. $\endgroup$ – mathreadler Aug 8 '17 at 16:49
  • $\begingroup$ @mathreadler Thanks for noticing, I will fix the problem with my notation. What I meant by $m$ was the eigenvector. And the integer on the loop is the number of that eigenvector. $\endgroup$ – Alireza Aug 8 '17 at 17:09
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    $\begingroup$ Yes that is what I guessed, but I just don't feel comfortable edit the content/meaning of other peoples questions. It's usually better if they do that themselves. $\endgroup$ – mathreadler Aug 8 '17 at 17:12
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    $\begingroup$ The Takagi factorization is almost what you want, except that we have the columns $m_i$ of $V$ are not quite eigenvectors and they satisfy $\langle m_i \mid m_j \rangle = \delta{ij}$. $\endgroup$ – Omnomnomnom Aug 8 '17 at 19:15

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