Normal matrices are of course useful to any linear algebra buff, not least because of the spectral theorem. However, taken as a whole, they tend to have some not-so-nice properties. For example:

  • unlike the unitary matrices, they are not closed under multiplication. That is, $A$ and $B$ being normal does not guarantee that $AB$ is normal
  • unlike the Hermitian matrices, they don't form a linear space. That is, $A$ and $B$ being normal does not guarantee that $A+B$ is normal

I'm wondering, then, if there are any nice properties of this set, or any good ways of thinking of how the set of normal matrices sits within $\Bbb C^{n\times n}$, or perhaps $\Bbb R^{n \times n}$.

Here's the way I see it: the normal matrices are the zero-set of the quadratic map $\phi:\Bbb F^{n \times n} \to \Bbb F^{n \times n}$ ($\Bbb F \in \Bbb{\{R,C\}}$) given by $$ \phi(A) = A^*A - AA^* $$ The set is closed. It is (I think) a smooth manifold in $\Bbb R^{n \times n}$, and it is connected in $\Bbb C^{n \times n}$.

Is it also connected in $\Bbb R^{n \times n}$? What is its dimension as a manifold in $\Bbb R^{n \times n}$? Do we know anything about its topological structure (Cohomology, for instance)? Is there some underlying algebraic structure to the "space" of normal matrices? Is there some interesting connection to draw between the unitary matrices and the Hermitian matrices that comes out of this analysis (the exponential map comes to mind)?

Any input here is appreciated. Thanks for reading :)

  • 3
    $\begingroup$ Are you sure this is smooth? It's conic (preserved under scaling), so being smooth I think implies that it is equal to a linear subspace, which you say is false. Perhaps its projectivization is smooth? $\endgroup$
    – user148177
    Sep 16 '14 at 3:28
  • $\begingroup$ One quick way to see the claimed fact is to use the Jacobian $J$ on the defining equations: conic means that they are homogeneous. Then, $J(0)$ is not full rank unless all the equations are also linear, and $0$ is a point on the variety (or space), so the variety is singular there. $\endgroup$
    – user148177
    Sep 16 '14 at 3:42
  • 1
    $\begingroup$ For dimension, perhaps one can use a group action. Every normal matrix is diagonalizable by a unitary matrix, so the unitary matrices act on normal matrices by conjugation. The biggest orbits have smallest stabilizers, i.e. normal matrices conjugate to a diagonal matrix with distinct eigenvalues. I think in fact unitary matrices act freely on this, so the dimension of such an orbit is the dimension of the unitary matrices. The parameter space for these orbits has real dimension $2n$, so it should be (for $\mathbb{C}$) $n^2 + 2n$. I'm no sure sure ab out this though. $\endgroup$
    – user148177
    Sep 16 '14 at 3:47
  • $\begingroup$ re: algebraic structure, it is an affine real algebraic variety in both cases, but not a complex variety (since taking conjugate transpose is not $\mathbb{C}$-linear). It is an intersection of quadrics and there might be something in the literature about that (maybe related to this? math.stackexchange.com/questions/261945/…) -- though I admit it's a stretch. Anyway, not something I know a whole lot about. $\endgroup$
    – user148177
    Sep 16 '14 at 3:56
  • $\begingroup$ @user148177 I was thinking "smooth" as meaning something like "differentiable", and so I forgot that the Jacobian has to be non-singular. Perhaps it is better to simply say that it is a conic. It is also a real variety. $\endgroup$ Sep 16 '14 at 3:56

Let me summarize my comments in an answer. I don't have answers to all the questions but I don't expect (myself) to really come up with any more, so here goes.

First, I am mostly thinking about this space as the algebraic variety over $\mathbb{R}$ (complex conjugation is not $\mathbb{C}$-scalar) cut out by the $n^2$ (in the $\mathbb{R}$ case) or $2n^2$ (in the $\mathbb{C}$ cae) equations given by $$A A^* - A^* A = 0.$$

This is a $\mathbb{R}$-conic variety in both cases, so in particular every point has a straight path to 0, so the space is path-connected in both cases.

Further, the only conic varieties which are smooth varieties (i.e. manifolds) are those cut out by linear equations. To see this, recall that the singular points are those where the Jacobian does not have full rank. Since the variety is cut out by homogeneous equations (i.e. it is conic) of non-linear degree, the Jacobian must vanish at zero. Further, zero is a point on the variety, so it is singular.

EDIT: Also, because it is a conic variety it is contractible, so homotopy equivalent to a point, so it has trivial topology. A more interesting question might be to ask about its projectivization, though I don't have much on that front.

Finally, one can compute the dimension by using the action of the unitary group on the normal matrices, at least in the case of $\mathbb{C}$. The special unitary group acts on the normal matrices by conjugation. Let us compute a "big" orbit, i.e. the orbit of a diagonal matrix with distinct diagonal entries. Only diagonal matrices can stabilize this point, and the diagonal matrices in the unitary group are isomorphic to $\mathbb{C}^*)^n$, i.e. have dimension $n$ , and so this orbit has dimension $\dim U(n) - n = n^2 - n$. Further, the parameter space for the orbits consists of parameterizing the diagonal entries; since the subset of this parameter space where the entries are distinct is a Zariski open locus, its dimension is equal to the parameter space, which is $2n$ (remember, real dimension). Thus the total space has dimension $n^2 + 2n - n = n^2 + n$. I mean dimension as a variety, meaning where the variety is smooth, it is locally of that dimension.

Hopefully all the above is correct.

EDIT: This MO post links to a few papers which have further answers: the normal variety is not orientable for $n \geq 2$, and is irreducible, etc.

  • $\begingroup$ Sounds reasonable to me. At the very least, the MO post looks like it has some interesting sources. Thanks! $\endgroup$ Sep 17 '14 at 17:35

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