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.
I have seen this question in Section 1.4, Exercise 6 of Linear Algebra by Hoffman & Kunze. I believe both the comment to the original question and the already-existing answer are assuming row-reduced is the same as row-reduced echelon.
In this text, a matrix is row-reduced if:
- the first non-zero entry in each non-zero row is equal to 1
- each column of the matrix which contains the leading nonzero entry of some row has all its other entries equal to zero.
With this definition, the three matrices are:
$ \begin{bmatrix} 0 & 0\\ 0& 0\end{bmatrix}, \begin{bmatrix} 1 & -1\\ 0& 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1& -1\end{bmatrix} $
Best Answer
Let
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \tag 1$
with
$AJ = JA; \tag 2$
writing
$AJ = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -b & a \\ -d & c \end{bmatrix} \tag 3$
and
$JA = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ -a & -b \end{bmatrix}, \tag 4$
we see in light of (2) that
$c = -b, \tag 5$
$d = a; \tag 6$
thus $A$ takes the form
$A = \begin{bmatrix} a & b \\ -b & a\end{bmatrix}; \tag 7$
we note that may write
$A = aI + bJ, \tag 8$
which evidently commutes with $J$; thus every matrix satisfying (2) is of the form (8). And under the correspondence
$i \longleftrightarrow J, \tag 9$
$A$ corresponds to the complex number $a + bi$.