Matrices for which $\mathbf{A}^{-1}=-\mathbf{A}$ Consider matrices $\mathbf{A}\in\mathbb{C}^{n\times n}$ (or maybe $A\in\mathbb{R}^{n\times n}$) for which $\mathbf{A}^{-1}=-\mathbf{A}$.
A conical example (and the only one I can come up with) would be $\mathbf{A} = \boldsymbol{i}\mathbf{I},\quad \boldsymbol i^2=-1$.
Now I have a few questions about this class of matrices:


*

*Are there more matrices than this example matrix (I guess yes) or can they even be generally constructed somehow?

*Are there also real matrices for which this holds?

*Now each matrix that is both skew-Hermitian and unitary fulfills this property. But does it also hold in the other direction, meaning is each matrix for which $\mathbf{A}^{-1}=-\mathbf{A}$ both skew-Hermitian and unitary (maybe this is simple to prove, but I don't know where to start at the moment, but of course I know if one holds the other has to hold, too)?

*Do such matrices have any practical meaning? For example I know that Hermitian and unitary matrices are reflections (in a general sense), but what about skew-Hermitian and unitary (if 3 holds)?


This is just for personal interrest without any practical application. I just stumbled accross this property by accident and want to know more about its implications and applications.
 A: Let $J=\left(\array{0 & 1 \\ -1 & 0}\right)$ as in Alexander's comment. Then $A=\operatorname{diag}(J,\dots,J)$ is a $2n \times 2n$ matrix with $A^2=-I$.
A: The equation $A^{-1} = -A$ is equivalent to $A^2 = -I$.  A matrix (over the complex numbers) that satisfies this must be similar to a diagonal matrix with diagonal entries $i$ and $-i$.  A real matrix that satisfies it must be similar to lhf's example. 
A: Note that the condition is invariant under conjugation, so any conjugate of a matrix satisfying this property also satisfies this property. 
It is simpler to write the condition as $A^2 = -I$. Note that taking determinants of both sides gives $(\det A)^2 = (-1)^n$, so if $n$ is odd then $A$ cannot be real. (Another way to see this is to note that, since $x^2 + 1$ is irreducible over $\mathbb{R}$, the characteristic polynomial of $A$ is necessarily $(x^2 + 1)^k$ for some $k$.) lhf's example shows that real examples always exist when $n$ is even.
Because the polynomial $x^2 = -1$ has no repeated roots, $A$ is diagonalizable with eigenvalues $\pm i$ and the converse holds. 
It is bad practice to ask whether a matrix is skew-Hermitian or unitary. This is not really a property of a matrix. It is a property of a linear operator on a complex inner product space. It should not be hard to construct examples which are neither skew-Hermitian nor unitary by conjugating a diagonal matrix with entries $\pm i$ by a non-unitary matrix. 
A: It is clear that your condition is equivalent to $-I=A^2$ (Multiply both sides with A). Now, Any matrix B that satisfy $B^2 = I$, will give a matrix $A = \pm i B$ that satisfy your relation, since $(\pm iB)^2 = i^2 B^2 = -I.$
Conversely, If $A^2 = -I$ then $B = \pm i A$ yields a matrix $B$
with the property $B^2 = 1.$
Thus, you are essentially looking for matrices that satisfy $B^2=I$
These matrices, if I am not mistake, are essentially reflections
of different types.
