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Hi everyone,

Ive been looking at the following problem, but its not entirely in my area and some potential solutions seem to rely on algebraic geometry. Maybe thats just a complicated way to solve the problem.

Im wondering if anyone can give me a good reference or answer to this question

For any set of $n×n$ matrices $A_1,A_2,...,A_k$, such that

  1. they share only a SINGLE eigenvector in common
  2. the joint commutant of $A_1,A_2,...,A_k$ and $A^*_1,A^*_2,...,A^*_k$ is trivial,

how can I guarantee there exists only this single one dimensional invariant subspace (corresponding to the span of that eigenvector)?

Googling around, it seems some progress has been made on saying whether invariant subspaces of dimension greater than 2 exist, but they rely on the fact that the matrices $A_1,A_2,...,A_k$ are chosen to have pairwise distinct eigenvalues.

Thanks for the help!

edit: changed some wording so I dont mislead anyone.

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    $\begingroup$ What is the "joint commutant"? $\endgroup$
    – user1688
    Commented Jun 19, 2013 at 11:14
  • $\begingroup$ Matrices with pairwise distinct eigenvalues are generic. $\endgroup$
    – Peter Michor
    Commented Jun 19, 2013 at 12:43
  • $\begingroup$ Oh, I mean that the only matrices which commutes with every matrix $A_1,...,A_k$ and their Hermitian conjugates $A^*_1,...,A^*_k$ are those which are scalar multiples of the identity. Sometimes this is denoted $\{A_1,...,A_k,A^*_1,...,A^*_k\}'=\mbox{span}(\mathbb{I})$. $\endgroup$
    – jeremy
    Commented Jun 19, 2013 at 12:45
  • $\begingroup$ Hm okay yea, I mean generic in the sense that I put no restrictions on the conditions of the eigenvalues. i.e. $A_1,..A_k\in\mathcal{M}^n(\mathbb{C})$ and can have degenerate eigenvalues. $\endgroup$
    – jeremy
    Commented Jun 19, 2013 at 12:49

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