So, I have a complex $4n \times 4n$ antisymmetric matrix, $A$ and it has a non-degenerate spectrum. The matrix $A$ then has eigenvalues given by

$$ \beta_{1}, -\beta_{1}, \beta_{2}, -\beta_{2}, ... , \beta_{2n}, -\beta_{2n} $$

where $\text{Re}(\beta_{j})\geq0$. $A$ has $4n$ eigenvectors $v_{1}, ... , v_{4n}$ with $ A v_{2j-1} = \beta_{j} v_{2j-1}$ and $ A v_{2j} = -\beta_{j} v_{2j}$.

Let $V$ be the matrix whose rows are the eigenvectors of $A$. Then it is possible to write $A$ as,

$A = V^{T} \Lambda V$

where $\Lambda = \text{diag}(\beta_{1},...,\beta_{2n}) \otimes \begin{pmatrix}0&1\\-1&0\end{pmatrix}$

In addition $V$ satisfies the property that $VV^{T} = I_{2n} \otimes \begin{pmatrix} 0&1\\1&0\end{pmatrix}$ ($I_{2n}$ is the $2n \times 2n$ identity matrix).

So Mathematica does not arrange the eigenvalues in the aforementioned pattern when I use Eigensystem and so $V$ does not satisfy the above property.

My question is: Is there an elegant way to order the eigenvalues in the way I mentioned so that $V$ has the above property?

Edit: Here is an example



m = ArrayFlatten[{{-Transpose[x],y},{zero,x}}];
u = 1/Sqrt[2.]*KroneckerProduct[{{1,-I},{1,I}},IdentityMatrix[2n]];
A = ConjugateTranspose[u].m.u;
A = 1/2*(A-Transpose[A]);

{val,vec} = Eigensystem[A];


{5.07121, -5.07121, 0.0978161 + 0.839127 I, -0.0978161 - 0.839127 I, 
 0.0978161 - 0.839127 I, -0.0978161 + 0.839127 I, 
 0.103329 + 0.682027 I, 
 0.103329 - 0.682027 I, -0.103329 + 0.682027 I, -0.103329 - 
  0.682027 I, -0.495647 - 0.195394 I, -0.495647 + 0.195394 I, 
 0.495647 + 0.195394 I, 
 0.495647 - 
  0.195394 I, 0.513024, -0.513024, -0.325981, 0.325981, -0.251215,

As can be seen, the eigenvalues are not in the aformentioned order.

  • 1
    $\begingroup$ One could use Ordering[] to sort the eigenvalues and eigenvectors in a desired manner. Do you have an example matrix? $\endgroup$ – J. M. will be back soon Sep 30 '18 at 2:36
  • $\begingroup$ I can give the matrix I am dealing with but the code is a little lengthy (~ 40 lines). I can tell you that the matrix is complex antisymmetric as I said and its eigenvalues come in pairs of 4, $\beta_{j}, -\beta_{j}, \beta_{j}^{*}, -\beta_{j}^{*}$ (again $\text{Re}(\beta_{j}) \geq 0$). I'm having a bit of trouble giving an example matrix that satisfies this property that is succinct to post. $\endgroup$ – user1058860 Sep 30 '18 at 7:56
  • $\begingroup$ I can speak only for myself, but without a concrete matrix I am so behind your state of information that I find it rather hopeless to successively mull over your problem... $\endgroup$ – Henrik Schumacher Sep 30 '18 at 10:04
  • $\begingroup$ An succinct example matrix has been posted @J.M.issomewhatokay $\endgroup$ – user1058860 Sep 30 '18 at 21:41
  • $\begingroup$ In your definition of $\Lambda$, the dimensions on the right-hand side don't match the left-hand side. Maybe you want a symplectic unit matrix there, and drop the negative eigenvalues? $\endgroup$ – Jens Sep 30 '18 at 22:30

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