1
$\begingroup$

Consider in the complex vector space $\mathbb{C}^n$. For an $n\times n$ complex matrix $A$. Is it true that $A$ has an invariant subspace of dimension $k$ ($k\le n$) if and only if both $A+A^*$ and $A-A^*$ have the same invariant subspace of dimension $k$? Here $A^*$ means the conjugate transpose of $A$.

Edited How to show that $A$ has only trivial invariant subspaces if and only if the only subspaces that are simultaneously under both $A+A^*$ and $A−A^*$ are the trivial subspaces?

  • 1
    Is $\mathcal{C}$ meant to represent the complex numbers? If so, you can get $\mathbb{C}$ with the command `\mathbb{C}` (`mathbb` stands for "math blackboard bold")2012-04-23
  • 0
    Your phrasing is unclear to me: "the only invariant subspace under both $A+A^*$ and $A-A^*$ are trivial" can mean: (i) the only $W$ that is *simultaneously* invariant under $A+A^*$ and under $A-A^*$ are $W=\{0\}$ and $W=\mathbb{C}^n$; or (ii) the only $W$ that are invariant under $A+A^*$ are $W=\{0\}$ and $W=\mathbb{C}^n$, and the only $Y$ that are invariant under $A-A^*$ are $Y=\{0\}$ and $Y=\mathbb{C}^n$. The latter is, at least a priori, stronger than the first. Which one do you mean?2012-04-23
  • 0
    I meant (i). Sorry for the confusion.2012-04-23

2 Answers 2