1
$\begingroup$

Consider the square matrix $M(k)$ of dimension 6

$ M(k) = \frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 0 & 1 & i & 0 & 0\\ 0 & 0 & 0 & 0 & ie^{-ik} & e^{-ik} \\ 0 & 0 & 0 & 0 & e^{i\frac{2\pi}{3}} & ie^{i\frac{2\pi}{3}} \\ ie^{i\frac{2\pi}{3}} & e^{i\frac{2\pi}{3}} & 0 & 0 & 0 & 0\\ e^{ik}e^{-i\frac{2\pi}{3}}& ie^{ik}e^{-i\frac{2\pi}{3}} & 0 & 0 & 0 & 0 \\ 0 & 0 & ie^{-i\frac{2\pi}{3}} & e^{-i\frac{2\pi}{3}} & 0 & 0 \\ \end{bmatrix}$

The code for this matrix is at the end of the question. One of the eigenvalues $\lambda(k)$ is of the form

$\lambda = \frac{ [\sqrt{2}cos(k)-\sqrt{-15+cos(2k)} ]^{1/3} }{ 2^{2/3} }$

The characteristic matrix $C$ of $M$ is defined as

$C(k) \equiv M(k) - I\lambda(k)$

The rank of $C$ should be less than its dimension because $Det[C]=0$, i.e. in this case the rank of $C$ should be less than 6. But, if you compute MatrixRank[$C$] in Mathematica, it returns 6. On the other hand, if you compute Det[$C$] and then fully simplify the expression, it gives you zero.

So there is clearly a problem here. Mathematica knows that the determinant of the matrix $C$ is zero, but it can't see that its rank is lower than its dimension.

As a consequence, Mathematica returns nothing when you ask it for the NullSpace of $C$. This is what I want, the NullSpace of $C$. Any advice on how to solve this?

By the way, if you compute MatrixRank[C] for a particular value of k, then Mathematica returns 5 (which is the correct answer). But for unspecified k, it gives 6 (which is the incorrect answer).

M = {{0,0,1/Sqrt[2],I/Sqrt[2],0,0},
 {0,0,0,0,(I E^(-I k))/Sqrt[2],E^(-I k)/Sqrt[2]},
 {0,0,0,0,E^((2 I π)/3)/Sqrt[2],(I E^((2 I π)/3))/Sqrt[2]},
 {(I E^((2 I π)/3))/Sqrt[2],E^((2 I π)/3)/Sqrt[2],0,0,0,0},
 {E^(I k-(2 I π)/3)/Sqrt[2],(I E^(I k-(2 I π)/3))/Sqrt[2],0,0,0,0},
 {0,0,(I E^(-((2 I π)/3)))/Sqrt[2],E^(-((2 I π)/3))/Sqrt[2],0,0}}

\[Lambda] = (Sqrt[2]Cos[k]-Sqrt[-15+ Cos[2k]])^(1/3)/(2^(2/3))
$\endgroup$
  • $\begingroup$ I think you are looking for the eigenvectors of M. You can find the eigenvectors corresponding to specific eigenvalues using Eigensystem. $\endgroup$ – mikado May 11 at 11:48
  • 1
    $\begingroup$ @mikado - If you plug in values of k (I tried k->1 and k->2) into the eigenvalues, one of them seems to equal the OPs lambda. $\endgroup$ – bill s May 11 at 13:23
  • $\begingroup$ @mikado, thanks for your suggestion. As you say, I am looking for the eigenvectors. But, for some reason, Eigensystem does not work properly for this matrix. The eigenvectors and eigenvalues it returns are discontinuous wrt k, when they shouldn't be. The \[Lambda] that I give should be correct. I got it by using CharacterisitcPolynomial, and then solving this polynomial. I didn't use Eigensystem because it doesn't work properly for this matrix $\endgroup$ – Ben Pepper May 11 at 14:39
2
$\begingroup$

You can pass in a stronger ZeroTest:

c = M - λ IdentityMatrix[6];

MatrixRank[c, ZeroTest -> PossibleZeroQ @* Simplify]
5

In fact there's a similar example here.

$\endgroup$
  • $\begingroup$ This works, thanks. But there's still a problem with getting the non-trivial solutions. LinearSolve[c, Table[0,{i,2m}], ZeroTest -> PossibleZeroQ @*Simplify] returns 0, but I want the non-trivial solution. Any advice for this? $\endgroup$ – Ben Pepper May 11 at 19:12
  • $\begingroup$ LinearSolve will only return one solution and instead you should use NullSpace with the custom ZeroTest. See here. $\endgroup$ – Chip Hurst May 11 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.