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An exercise on the manifold of derogatory matrices

17 Novembre 2013, 18:21pm

Publié par Fabien Besnard

Here is a nice little exercise. Let us consider the set H_N of all hermitian NxN matrices. In this circumstance, a matrix M is called derogatory if it has at least one multiple eigenvalue (in more general circumstances the definition is more complex, but never mind). In fact we are interested in the set A of hermitian matrices which are block diagonal, with the size (and place) of the blocks fixed. In other words we are looking at the direct sum A of H_{n_1},...,H_{n_k} with n_1+...+n_k=N.

 

We suppose that k>1.

 

Let us call V the manifold of derogatory matrices inside A. It decomposes into different components, according to whether the multiplicities fall inside the same block or not. We are interested in the highest dimensional components, which have dimension N-1 (why ?) and correspond to the case where the multiplicities are across different blocks (say the first two for concreteness).

 

The question is : what is the equation of the tangent hyperplane to V at some matrix a the two first blocks of which share an eigenvalue ?

 

I solved this in a rather heuristic way, guessing the answer from a simple example, and then checking that it holds generally. I wonder if there is a more systematic ( but still simple) way to derive the result.

 

I'll give the answer at the end of the week.

 

 

Update 20/11/2013 : Since nobody gave the answer, here is an indication. You might like to first work out the case where the blocks have all size 1.

Commenter cet article

Fabien Besnard 25/11/2013 17:53


Let a=U(x p+...+x q+...)U* where U is a unitary matrix and the dots stand for a linear combination of eigenprojectors p_1,...,p_m (m=N-2). The tangent vectors to V at a come in two different
kinds : those which are obtained by an infinitesimal variation of U, and are Frobenius-orthogonal to all the eigenprojectors of a, hence to p-q, and those which come from an infinitesimal
variation of the eigenvalues, keeping U fixed. These guys are combinations of p+q, and the p_i's. Hence they are all orthogonal to p-q.


The equation which we seek is thus Tr((p-q)(a-m))=0, where m belongs to A.

Fabien 23/11/2013 17:13


As promised here is the answer. We look for a matrix which is orthogonal to V at a with respect to the Frobenius scala r product.  As can be guessed from the case where all blocks have size
one, p-q does the job, where p and q are the eigenprojections  corresponding  to the shared eigenvalue respectively for the first and second block. That p-q is orthogonal to V can be
proven in the following way (to be continued)