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Let $$ \mathcal{J}=\{A\in M_n(\mathbb{C}):\ A \text{ is a Jordan matrix}\} $$ Then it is well-known that the similary orbit of $\mathcal{J}$ is all of $M_n(\mathbb{C})$.

The question is,

what is the unitary orbit of $\mathcal{J}$? Is it dense?

It cannot be all of $M_n(\mathbb{C})$, because every matrix in $\mathcal{J}$ an its unitary conjugates have the property that eigenvectors corresponding to different eigenvalues are orthogonal to each other.

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up vote 4 down vote accepted

It cannot be dense except in the trivial case of $n=1$. The (real) dimension of $U(n)$ is $n^2$ while the dimension of $Gl_n(\mathbb{C})$ is $2n^2$ Since $\mathcal{J}$ has dimension $n$ (in the sense that it's a union of the diagonal matrices together with unions of various Jordan blocks things of smaller dimension since we get to choose fewer eigenvalues), the $U(n)$ orbit through $\mathcal{J}$ has dimension at most $n^2+n$.

But $n^2 + n\leq 2n^2$ unless $n=n^2$, i.e., $n=0$ or $n=1$.

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Actually $\cal J$ should have real dimension $2n$ since the eigenvalues can be complex. Of course that doesn't really affect the argument. – Robert Israel Mar 9 '12 at 4:43
You're right - that was silly of me. Although when $n=2$, it leaves something to prove. – Jason DeVito Mar 9 '12 at 4:45
A slightly easier approach is to notice that for compact Lie groups $G$, the orbit through a closed set is still closed. Since $\mathcal{J}$ is closed, the $U(n)$ orbit through it will still be closed. Hence, it will be dense iff it's the whole space (and it's easy to see the orbit is not the whole space). – Jason DeVito Mar 11 '12 at 18:38

It is the set of $n \times n$ matrices $A$ such that there is an orthormal basis $\{u_j:\ j=1\ldots n\}$ that is the union of blocks $\{u_j:\ a \le j \le b\}$ where $u_a$ is an eigenvector of $A$ for some eigenvalue $\lambda$ and $A u_j = \lambda u_j + u_{j-1}$ for $a+1 \le j \le b$. Well, that's just restating the fact that the matrix for $A$ in this basis should be in Jordan form.

EDIT: here's a necessary condition, I'm not sure it's sufficient. $A = B + N$ where $B$ is a normal matrix and $N$ a nilpotent matrix that commutes with $B$, and such that $N^k (N^k)^*$ and $(N^k)^* N^k$ are orthogonal projections for each positive integer $k$.

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I like that. I'm interested however on information about the set itself, and not that much on its elements. For instance, is it dense? I'll add that question. – Martin Argerami Mar 8 '12 at 23:18

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