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What can one say about the set of all $n$-dimensional square matrices $A \in \text{GL}_n(\mathbb{C})$ that have an inverse with entries out of $\mathbb{C}$ with the properties:

  • unitary $:\Leftrightarrow A^*= A^{-1}$
  • hermitian $:\Leftrightarrow A^* = A$

where $A^*$ is the conjugate transpose of $A$.

What obviously follows is $A^{-1} = A$ The most simple matrix that is in this set is the identity matrix. Are there others? How do they look like?

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    The map $A\mapsto(A+I)/2$ is a bijection from the set of self-adjoint unitary matrices to the set of self-adjoint projections. This is a straightforward computation that carries over to operators on an arbitrary Hilbert space $H$, where $P\mapsto P(H)$ is a bijection from the set of self-adjoint projections on $H$ to the set of closed subspaces of $H$.2011-01-07

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Such a matrix is such that $A^2=I$, so it is diagonalizable, and its possible eigenvalues are $+1$ and $-1$. Since it is unitary, the eigenspaces corresponding to $1$ and to $-1$ are orthogonal.

Conversely, every diagonalizable matrix with eigenvalues contained in $\{+1,-1\}$ and orthogonal eigenspaces is of that form.

It follows that the set of your matrices is in bijection with the set of subspaces of $\mathbb C^n$. Explicitely: If $V$ is one such subspace, there is a unique linear transformation $f:\mathbb C^n\to\mathbb C^n$ such that $V$ and $V^\perp$ are eigenspaces for the eigenvalues $1$ and $-1$. The matrix $A_V$ of $f$ with respect to the standard basis of $\mathbb C^n$ satisfies your conditions. The bijection is $V\in\{\mathrm{subspaces\;of\;}\mathbb C^n\}\leftrightarrow A_V.$

As a consequence, considered as a whole, your set is a disjoint union of submanifolds homeomorphic to Grassmannian varieties. Literally books have been written about them.

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    +1 Sounds creepy, but quite interesting. As often in maths.2011-01-06
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(in addition to Mariano's answer), these are sometimes called Householder transformations.

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    @MAriano: Orthogonal reflection? How can that possibly become a Household name? ;-)2011-01-06
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A simple answer for $A \in \mathbb{R}^{2 \times 2}$:

Yes, there are more matrices. They are of the form

$A = \begin{pmatrix}\cos \alpha & \sin \alpha\\ \sin \alpha & -\cos \alpha \end{pmatrix}$.

It is obvious that these matrices are symmetric.

The inverse is $\frac{1}{(\cos \alpha) \cdot (-\cos \alpha) - sin^2 \alpha} \begin{pmatrix}-\cos \alpha & -\sin \alpha\\ -\sin \alpha & \cos \alpha \end{pmatrix} = \frac{-1}{\cos^2 \alpha + sin^2 \alpha} \begin{pmatrix}-\cos \alpha & -\sin \alpha\\ -\sin \alpha & \cos \alpha \end{pmatrix} = A$

Maybe you can find such forms for $\mathbb{C}^{n \times n}$, too.

I know this is not exactly an answer to your question. But I think this explicit, simple form is a nice answer for a part of the question.

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    You're welcome! Oh, the signs are still missing in the previous step! :)2012-09-12