I've heard a couple of times some people say that in a way, Hermitian matrices are to matrices as real numbers are to complex numbers. I know two examples where this is sort of true:

Complex conjugate vs conjugate transpose We can realize complex numbers as a set of matrices, where complex number $z = a + bi$ corresponds to a matrix $$A = \begin{bmatrix}a & -b \\ b & a\end{bmatrix}$$

Then the conjugate $\bar{z} = a - bi$ corresponds to the conjugate transpose $A^*$. A complex number is real if and only if $z = \bar{z}$. On the other hand, matrix $A$ is called Hermitian if and only if $A = A^*$.

Polar form vs polar decomposition: We can represent every complex number $z$ as $z = re^{i\varphi}$, where $r \geq 0$ and $\varphi \in [0, 2\pi[$. This representation is unique when $z \neq 0$. On the other hand, any $n \times n$ matrix $A$ with complex entries can be represented as $A = RU$ where $R$ is positive semidefinite (thus Hermitian) and $U$ is unitary. This representation is unique when $A$ is invertible. Also, if $\det A = re^{i\varphi}$, then $\det R = r$ and $\det U = e^{i\varphi}$.

Taking the second example really far, you could say positive semidefinite matrices are like nonnegative real numbers and unitary matrices are like points on the unit circle.

Are there more examples? I think this is interesting, but is thinking like this useful at all? It seems like oversimplifying things and I doubt if this leads to anything other than some fun facts.


1 Answer 1


Well, one trivial connection is that if you look at $1\times 1$ matrices (which have only a single complex entry), then you'll find that it is real iff it is Hermitian, its complex conjugate is its conjugate transpose, and its polar decomposition is the polar form.

Also, just like a complex number can be uniquely decomposed into a real and an imaginary part ($z = a+\mathrm ib$ with real $a,b$), a complex matrix can be uniquely decomposed into a Hermitian and an "anti-Hermitian" part, i.e,. $M =A + \mathrm iB$ with $A$ and $B$ Hermitian. And just like $\Re(z)=\frac12(z+\bar z)$ and $\Im(z)=\frac1{2\mathrm i}(z-\bar z)$, the Hermitian part of a matrix is $\frac12(M+M^*)$ and the "anti-Hermitian" part is $\frac1{2\mathrm i}(M-M^*)$.

Moreover, just like $\bar zz$ is a non-negative real number, $M^*M$ is a positive semidefinite matrix.

Another point: Hermitian matrices have real eigenvalues, and unitary matrices have eigenvalues of the form $\mathrm e^{\mathrm i\phi}$.

About the usefulness of the analogy:
In classical physics, observables should be real. In quantum physics, observables are represented by Hermitian matrices. Also, the quantum analogue to probability densities, which are non-negative functions with integral $1$, are density operators, which are positive semidefinite matrices with trace $1$. So there's indeed some connection.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.