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?

$\endgroup$
3
  • 1
    $\begingroup$ 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") $\endgroup$
    – Arturo Magidin
    Apr 23, 2012 at 18:41
  • $\begingroup$ 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? $\endgroup$
    – Arturo Magidin
    Apr 23, 2012 at 20:41
  • $\begingroup$ I meant (i). Sorry for the confusion. $\endgroup$
    – Sunni
    Apr 23, 2012 at 21:00

2 Answers 2

Reset to default
2
$\begingroup$

No. For example, for $A = \pmatrix{1 & 0\cr 1 & 1\cr}$, the only $1$-dimensional invariant subspace is spanned by $\pmatrix{0\cr 1\cr}$, but the $1$-dimensional invariant subspaces of $A + A^* = \pmatrix{2 & 1\cr 1 & 2\cr}$ are the spans of $\pmatrix{1\cr -1\cr}$ and $\pmatrix{1\cr 1\cr}$, while those of $A - A^* = \pmatrix{0 & -1\cr 1 & 0\cr}$ are the spans of $\pmatrix{i\cr 1\cr}$ and $\pmatrix{-i\cr 1\cr}$.

$\endgroup$
2
  • $\begingroup$ How to show that $A$ has only trivial invariant subspaces if and only if the invariant subspace under both $A+A^*$ and $A−A^*$ are trivial? $\endgroup$
    – Sunni
    Apr 23, 2012 at 19:25
  • $\begingroup$ There are no $n \times n$ matrices (with $n \ge 2$) over an algebraically closed field that have only trivial invariant subspaces, if by "trivial" you mean dimension $0$ or $n$. Every such matrix has at least one eigenvector. $\endgroup$
    – Robert Israel
    Apr 23, 2012 at 20:36
1
$\begingroup$

If there is any matrix $A$ without an invariant subspace of a certain dimension $k$ (maybe this is easy, but I couldn't quickly come up with an example), then your assertion cannot be true: because $A+A^*$ is selfadjoint and so it has invariant subspaces of every dimension. Similarly, $i(A-A^*)$ is also selfadjoint and so again it has invariant subspaces of every dimension, which implies that so does $A-A^*$.

$\endgroup$
2
  • $\begingroup$ Sorry, I did not express what I really mean. Please allow me to correct the question. $\endgroup$
    – Sunni
    Apr 23, 2012 at 18:53
  • $\begingroup$ Using Jordan canonical form, an $n \times n$ matrix over an algebraically closed field always has at least one invariant subspace of any dimension $\le k$. $\endgroup$
    – Robert Israel
    Apr 23, 2012 at 19:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .