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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:

  1. Are there more matrices than this example matrix (I guess yes) or can they even be generally constructed somehow?
  2. Are there also real matrices for which this holds?
  3. 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)?
  4. 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.

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4  
$\left(\array{0 & 1 \\ -1 & 0}\right)$ –  Alexander Thumm Dec 8 '11 at 20:41
    
@AlexanderThumm Of course, as simple as it can be, me stupid! But what about more general matrices? –  Christian Rau Dec 8 '11 at 20:44
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Easier to deal with $A^2=-I$, so you can say something about the eigenvalues of $A$. –  Thomas Andrews Dec 8 '11 at 20:45
    
Considering Characteristic equation, your matrix $A$ must satisfy $$cA+dI=0 $$ where $c,d\in \mathbf{C}$ and we know their expression in terms of sum of K-ordered minors of $A$. –  Tapu Dec 8 '11 at 21:21

4 Answers 4

up vote 4 down vote accepted

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.

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Thanks for your answer, but I have difficulties to understand the last sentence. What do you mean with "by conjugating by a non-unitary matrix". Now I understand that it is invariant under conjugation. But aren't the skew-Hermitian and unitary properties invariant under conjugation, too? I don't know how cojugation helps with constructing a counter-example. Or do you mean something different (I don't really understand what conjugation by a non-unitary matrix means)? –  Christian Rau Dec 9 '11 at 1:21
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@Christian: skew-Hermitian and unitary are only invariant under conjugation by a unitary matrix. "By" refers to the matrix $G$ doing the conjugating $D \mapsto GDG^{-1}$. –  Qiaochu Yuan Dec 9 '11 at 1:22
    
I just cannot get what conjugation by a unitary matrix means. I guess with conjugation you mean not just the simple complex conjugate of the matrix elements? Sorry, but I'm not that deeply versed in linear algebra, especially if it gets too theoretical. –  Christian Rau Dec 9 '11 at 1:24
    
Ah, ok. Now I understand what you mean by conjugation. And I understand why this keeps the mentioned property but destroys the unitarity property. Thanks again. –  Christian Rau Dec 9 '11 at 1:26

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 $A^2=-I$.

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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.

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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.

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The standard $\mathbb R^{2\times 2}$ example is not a reflection, but a rotation by 90°. In fact, reflections cannot be examples, as they always satisfy $B^2=I$ rather than $B^2=-I$ –  Henning Makholm Dec 8 '11 at 21:09
    
@HenningMakholm He just means that a matrix with the searched property can be constructed from any reflection matrix. Though his last paragraph is a bit obsolete, but +1 for the first paragraph. –  Christian Rau Dec 9 '11 at 1:56

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