Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I am using Mathematica to analyze a real, self-adjoint matrix $H$ of the size $32 \times 32$, which comes from a physics problem. In the picture there is also a matrix $Q$ which commutes with $H$.

I would like to see what $Q$ looks like in the eigenbasis of $H$, and I run into some technical issues that I was hoping someone might be kind enough to resolve.

To find an eigenbasis of $H$ is not difficult - the functions Eigenvalues and Eigenvectors work well enough. The next step is very simple in theory: combine the eigenvectors into a transition matrix $P$, and compute the matrix of $Q$ relative to the new basis as $Q' = P^{-1}Q P$. In practice, it seems that the computation of the inverse requires a lot of computation, which is time consuming now, and will probably become infeasible for larger $H$. After a moment's thought, I realized that the matters could greatly helped by the fact that $H$ is self-adjoint: it follows that it has an orthonormal eigenbasis, for which computing the inverse amounts to computing the adjoint. This would solve the matters if $H$ had all eigenvalues distinct (then $P$ mentioned above would already be self-adjoint, maybe up to re-scaling the rows), but this is not the case here.

Hence the question: Given a self-adjoint matrix $H$, together with its eigenvectors, assuming that the eigenspaces are more than $1$-dimensional, how can I effectively compute the inverse matrix? (Alternatively: how else can I find the matrix form of $Q$ relative to the eigenbasis of $H$?)

share|improve this question
3  
This has also been addressed in the comments and answers to Can Eigenvalues and Eigenvectors be assumed to return the same ordering? –  Jens Apr 29 '13 at 17:17
1  
QRDecomposition might be useful for obtaining an orthogonal change of basis. –  Daniel Lichtblau Apr 29 '13 at 19:11
add comment

1 Answer 1

up vote 9 down vote accepted

As $P$ is explicitly constructed from eigenvectors of a self-adjoint matrix, it is unitary, i.e $P P^\dagger = I\qquad$ where the $\dagger$ is the conjugate transpose (or Hermitian conjugate, if you prefer). So, calculating the inverse is simply ConjugateTranspose[P] which is much faster than calculating it using Inverse. That said, you have to ensure that the eigenvectors are orthonormal to properly form $P$ as Eigenvectors only guarantees linear independence in degenerate eigenspaces. A simple example of this effect is

Eigensystem[#.DiagonalMatrix[{-1, 1, 1}].ConjugateTranspose[#]]& @
  RotationMatrix[2 Pi/3, {1, 0, 1}]

which gives

{{-1, 1, 1}, {{1/3, Sqrt[2/3], 1}, {-3, 0, 1}, {-Sqrt[6], 1, 0}}}

As you can see, in this case the eigenvectors are not normalized, nor are the eigenvectors for $\lambda = 1$ orthogonal to each other. In this state, $P$ is not unitary:

#.ConjugateTranspose[#]& @ %[[2]]
(* {{16/9, 0, 0}, {0, 10, 3 Sqrt[6]}, {0, 3 Sqrt[6], 7}} *)

This must be corrected to be useful, and it can be done using Orthogonalize. The code is as follows:

P = Orthogonalize@Eigenvectors@H;
Qprime = P.Q.ConjugateTranspose[P];

Note, this uses $P Q P^\dagger$, instead of $P^\dagger Q P$ as Eigenvectors explicitly generates a matrix where the rows are the eigenvectors of $H$.

share|improve this answer
    
Thank you, that was just what I needed. And yes, I did mean $Q' = P^{-1} Q P$; just corrected it for increased readibility. –  Feanor Apr 29 '13 at 16:26
    
You're welcome. I removed the question from my answer. Incidentally, I have been bitten by the linear independence issue a number of times, and it is easy to miss being a single line under the Details section of the documentation. –  rcollyer Apr 29 '13 at 16:30
    
+1. But didn't you mean to write "between nondegenerate" instead of "in degenerate"? –  whuber Apr 29 '13 at 16:47
    
@whuber No, I meant "within a degenerate eigenspace," which may not be the correct term in this case. Probably, this is more correct: "between the eigenvectors for a degenerate eigenvalue." Does that work better? –  rcollyer Apr 29 '13 at 16:55
    
I'm not sure. Because eigenvectors of a Hermitian matrix with different eigenvalues are guaranteed orthogonal, the only issue concerns within degenerate eigenspaces. The help asserts that orthonormal eigenvectors for degenerate eigenspaces are returned. That leaves me wondering why Orthogonalize is needed at all. But I'm not really sure what you're saying because of the sudden mysterious appearance of "$U$" in your answer. –  whuber Apr 29 '13 at 17:54
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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