# Orthogonal and symmetric Matrices

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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What is a quadratic matrix? Is $A^H$ the conjugate transpose of $A$? –  Mariano Suárez-Alvarez Jan 6 '11 at 20:32
Quadratic means $n \times n$: $n$ rows & $n$ columns; $A^H$ means the conjugate transpose. –  a1337q Jan 6 '11 at 20:36
Strange. Most people would say square :) –  Mariano Suárez-Alvarez Jan 6 '11 at 20:40
Sorry, I had this course in german ;) –  a1337q Jan 6 '11 at 20:41
@a1337q: $A^*$ is more common than $A^H$ for conjugate transpose (though there seems to be little by way of very generally accepted notation). –  Arturo Magidin Jan 6 '11 at 20:44

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. –  a1337q Jan 6 '11 at 21:00

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+1 bc I know these but were actually unable to make a link (d'oh) –  a1337q Jan 6 '11 at 20:54
Only when the multiplicity of $-1$ is exactly $1$. In any case, I never understood why people introduce a new name for what's already called an orthogonal reflection! –  Mariano Suárez-Alvarez Jan 6 '11 at 20:55
@MAriano: Orthogonal reflection? How can that possibly become a Household name? ;-) –  Aryabhata Jan 6 '11 at 21:20

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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There's a small typo: you forgot the minus signs before the sines, while inverting $A$. –  Andrea Orta Sep 12 '12 at 10:28
Thanks for mentioning it. –  moose Sep 12 '12 at 11:27
You're welcome! Oh, the signs are still missing in the previous step! :) –  Andrea Orta Sep 12 '12 at 12:47