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Matrices such as

$$ \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} \text{ or } \begin{bmatrix} \cos\theta & i\sin\theta \\ -i\sin\theta & -\cos\theta \end{bmatrix} \text{ or } \begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{bmatrix} $$

are both unitary and Hermitian (for $0 \le \theta \le 2\pi$). I call the latter type trivial, since its columns equal to plus/minus columns of the identity matrix.

Do such matrices have any significance (in theory or practice)?

In the answer to this question, it is said that "for every Hilbert space except $\mathbb{C}^2$, a unitary matrix cannot be Hermitian and vice versa." It was commented that identity matrices are always both unitary and Hermitian, and so this rule is not true. In fact, all trivial matrices (as defined above) have this property. Moreover, matrices such as

$$ \begin{bmatrix} \sqrt {0.5} & 0 & \sqrt {0.5} \\ 0 & 1 & 0 \\ \sqrt {0.5} & 0 & -\sqrt {0.5} \end{bmatrix} $$

are both unitary and Hermitian.

So, the general rule in the aforementioned question seems to be pointless.

It seems that, for any $n > 1$, infinitely many matrices over the Hilbert space $\mathbb{C}^n$ are simultaneously unitary and Hermitian, right?

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The second example is neither unitary nor Hermitian. – Jonas Meyer Aug 13 '11 at 0:59
For your last question, note that the case $n=2$ implies the rest, because if $A$ is unitary and Hermitian then so is $\begin{bmatrix}A&0\\0&1\end{bmatrix}$. Of course, there are only $2$ when $n=1$. – Jonas Meyer Aug 13 '11 at 1:02
I don't know if this qualifies as significant, but $A\mapsto (A+I)\mathbb{C}^n$ is a bijection from the set of Hermitian unitaries to the set of subspaces of $\mathbb C^n$. See also… – Jonas Meyer Aug 13 '11 at 1:09
@Jonas: Thanks. I corrected the second example, and changed the condition to $n>1$. – M.S. Dousti Aug 13 '11 at 1:28
Thanks, but I mean $\begin{bmatrix} ai & \pm\sqrt{1+a^2} \\ \pm\sqrt{1+a^2} & -ai \end{bmatrix}$ is neither unitary nor Hermitian. – Jonas Meyer Aug 13 '11 at 1:33

6 Answers 6

up vote 14 down vote accepted

Unitary matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are on the unit circle. Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are real. So unitary Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $\pm 1$.

This is a very strong condition. As George Lowther says, any such matrix $M$ has the property that $P = \frac{M+1}{2}$ admits a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $0, 1$; thus $P$ is a Hermitian idempotent, or as George Lowther says an orthogonal projection. Of course such matrices are interesting and appear naturally in mathematics, but it seems to me that in general it's more natural to start from the idempotence condition.

I suppose one could say that Hermitian unitary matrices precisely describe unitary representations of the cyclic group $C_2$, but from this perspective the fact that such matrices happen to be Hermitian is an accident coming from the fact that $2$ is too small.

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A matrix $M$ is unitary and Hermitian if and only if $M=2P-1$ for an orthogonal projection $P$. That is, $P$ is Hermitian and $P^2=P$.

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Ha, yes. Of course orthogonal projections are quite interesting. Let me rephrase my answer. – Qiaochu Yuan Aug 13 '11 at 1:49
Or another way of saying it: complex Householder matrices are unitary and Hermitian. – J. M. Aug 13 '11 at 7:13
@George Lowther This is a side comment really but what is special about linear transformations $P$ such that $P = P^2$? – user38268 Aug 15 '11 at 13:26
@D Lim: Lots of things, which I don't have time to go into now. But, they are in one to one correspondence with closed subspaces. All hermitian operators can be built out of commuting sets of orthogonal projections (spectral decomposition) and, similarly, they are the basic building blocks of quantum mechanical observables (representing binary true-false information). – George Lowther Aug 15 '11 at 13:43

"Do such matrices have any significance (in theory or practice)?"

Yes, they certainly do. As I commented in George's answer, complex Householder matrices (a.k.a. elementary reflectors) are both unitary and Hermitian. In general, one can easily construct a Householder matrix $\mathbf H=\mathbf I-2\mathbf u\mathbf u^\dagger,\quad \|\mathbf u\|_2=1$ such that $\mathbf H\cdot\mathbf v=c\mathbf e_1$, where $\mathbf v$ is an arbitrary complex vector, $\mathbf e_1$ is the first column of the identity matrix, and $c$ is real. One can thus consider complex versions of the usual linear algebra algorithms that rely on orthogonal matrices, e.g. QR, SVD, Schur decompositions...

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Since no-one else seems to have said it (explicitly at least, although elements of order $2$ and projections are closely linked, as indicated in some answers), a unitary matrix which is also Hermitian is just a unitary matrix of multiplicative order at most $2$ (or, equivalently, a Hermitian matrix of multiplicative order at most $2$). For a matrix $A$ is unitary if an only if $A^{*} = A^{-1},$ where $*$ denotes "transposed conjugate", while $A$ is Hermitian if and only if $A^{*} = A.$ Hence if $A$ is both unitary and Hermitian, we have $A = A^{-1}$ (and $A$ is unitary). As for theoretical uses, the group ${\rm SU}_{n}^{\pm}(\mathbb{C})$ is generated by such matrices for every $n$, where ${\rm SU}_{n}^{\pm}(\mathbb{C})$ denotes the group of unitary $n \times n$ matrices of determinant $\pm 1$. This is clear for $n = 1$, and follows easily by induction, using the fact that ${\rm PSU}(n,\mathbb{C})$ is a simple group for $n > 1.$

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+1. Among the answers given so far, this one is my favorite! (I find the other answers very nice, but this one outstanding!) – Pierre-Yves Gaillard Aug 13 '11 at 10:23
Dear Geoff, you wrote $=A^{-1}$ instead of $A=A^{-1}$. – Pierre-Yves Gaillard Aug 13 '11 at 10:53
@Pierre@Thanks on both counts. I have corrected the typo. – Geoff Robinson Aug 13 '11 at 10:55

Yes, operators in quantum mechanics should preserve the norm of the physical state so they should be unitary. Also, eigenvalues of the operators should be real numbers so they should be Hermitian.

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Such matrices are used in the field of Quantum Computing. States of a Quantum system exist in a Hilbert space of dimension n. The space of operators or "gates" having affect on this Hilbert state space is the space of all n*n dimension Unitary matrices.

Vectors in this operator space that are Hermitian are extremely useful because they denote gates that are reversible (note how if A is Hermitian A=A^-1, so not only is the Quantum operation A invertible, it is in fact it's own inverse.)

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