I encountered the following and I am puzzled!... where am I wrong?

basic definition of hermitian matrix $H$ is $H=H^*$ if I take the outer product of the following complex vector V and its conjugate I get this

$V=\begin{bmatrix}{\rm e}^{j d_1},& {\rm e}^{j d_2},& {\rm e}^{j d_3}, &\ldots,& {\rm e}^{j d_N}\end{bmatrix}^{\rm T} $

$V^*=\begin{bmatrix}{\rm e}^{-j d_1},& {\rm e}^{-j d_2},& {\rm e}^{-j d_3}, &\ldots,& {\rm e}^{-j d_N}\end{bmatrix} $


$$H = V V^* = \begin{bmatrix} 1 & {\rm e}^{j (d_1 - d_2)} & {\rm e}^{j (d_1 - d_3)} & \ldots & {\rm e}^{j (d_1 - d_N)} \\ {\rm e}^{j (d_2 - d_1)} & 1 & {\rm e}^{j (d_2 - d_3)} & \ldots & {\rm e}^{j (d_2 - d_N)} \\ {\rm e}^{j (d_3 - d_1)} & {\rm e}^{j (d_3 - d_2)} & 1 & \ldots & {\rm e}^{j (d_3 - d_N)} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {\rm e}^{j (d_N - d_1)} & {\rm e}^{j (d_N - d_2)} & {\rm e}^{j (d_N - d_3)} & \ldots & 1 \end{bmatrix}$$

Now testing $H$ it is a hermitian, $H=H^*$.

However its determinant is zero, i.e. it is singular.

Testing for eigenvalues, we get one eigenvalue as zero and the rest is not zero.

For any $N$ the charcteristic equation has the following form regardless of changes in the values of variables $d_n$.


[(1-lambda)^N]-1= zero

this equation results in a one zero eigenvalue and the rest are both real eigen values and pairs of complex conjugates...if these add...the trace is always real value since the complex conjugates when added produce real values

I am confused... I am having complex eigen values for a hermitian matrix that satisfies all the definition properties?? kindly help...with refernces please

  • $\begingroup$ How did you compute the eigenvalues? It seems something must have gone wrong there. Your matrix is an outer product of two vectors, hence it is rank one. Therefore, it has one eigenvalue that is non-zero and all others are zero. The non-zero one should be equal to $N$. $\endgroup$
    – Florian
    Commented Dec 23, 2016 at 14:23
  • $\begingroup$ I computed my eigen valuesby getting the charcteristic equation as determinant of ( H-lambda*Ìdentity matrix)=0. it gives me along the diagonal (1-lambda)...when computing the determinant by the conventional method... diagonals result in (1-lambda)^N....other generated values are ones (N-1) minus N see the generated equation above...I tested this result manually and is willing to see if your test gave different results $\endgroup$ Commented Dec 23, 2016 at 14:54
  • $\begingroup$ Please don't send me files. ;-) I think there is no need to. What software are you using? In Matlab, it's these two lines: v = exp(jd(:)); disp(eig(vv')); $\endgroup$
    – Florian
    Commented Dec 23, 2016 at 15:11

1 Answer 1


Your matrix $\mathbf H$ is equal to $\mathbf H = \mathbf v \mathbf v^*$. Therefore, it is easy to see that it has one eigenvalue equal to $\left\|\mathbf v\right\|^2 = N$, since: $$\mathbf H \cdot \mathbf v = \mathbf v \cdot \mathbf v^* \cdot \mathbf v = \|\mathbf v\|^2\cdot\mathbf v.$$

All other eigenvalues are zero since your matrix is rank one.

If you are getting complex eigenvalues then very likely something has gone wrong in their computation.

  • $\begingroup$ I am surprised and puzzled because I am aware of this... $\endgroup$ Commented Dec 23, 2016 at 14:55
  • $\begingroup$ thanks...it is ckear now (I do things the old way...leaving computer program to last steps :-)...) $\endgroup$ Commented Dec 23, 2016 at 16:12

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