Are there matrices that satisfy these two conditions? That is, a matrix $A$ such that


What I know is that a skew-symmetric matrix with $n$ dimensions is singular when $n$ is odd.

  • $\begingroup$ Can you find one of size 2? Using only zeros, 1 and -1. $\endgroup$ – Mariano Suárez-Álvarez Aug 28 '17 at 3:59
  • $\begingroup$ Since you have $\mathbf A^\top\mathbf A=\mathbf I$ and $\mathbf A^\top=-\mathbf A$, you get $-\mathbf A^2=\mathbf I$. What do you think that matrix looks like? $\endgroup$ – J. M. is a poor mathematician Aug 28 '17 at 4:00
  • $\begingroup$ For the complex case (i.e., $A\in GL_n(\mathbb{C})$ with $A^\dagger = A^{-1} = -A$), just take a suitable diagonal matrix. To get to the real case from that one, use the embedding $GL_n(\mathbb{C}) \to GL_{2n}(\mathbb{R})$ given by the action of $\mathbb{C}$ on itself as a real vector space. $\endgroup$ – anomaly Aug 28 '17 at 4:07

The matrix $$A=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$$ is skew-symmetric and orthogonal. In even dimensions, we can always construct a skew-symmetric and orthogonal matrix as the direct sum of multiple copies of $A$. i.e. the matrix $$\bigoplus_{i=1}^k A=\underbrace{\begin{bmatrix} A &&&\\ &A&&\\ &&\ddots & \\ &&&A\\ \end{bmatrix}}_{k \text{ copies}}$$ is a $2k \times 2k$ orthogonal and skew-symmetric matrix

In odd dimensions however, there is no real matrices which are skew-symmetric and orthogonal. As you already know, real skew-symmetric matrices are singular in odd dimensions so they must have at least one eigenvalue which is zero. Therefore if $B$ is an $n\times n$ matrix with $n$ odd, it must be the case that $\text{det}(B)=0$. If $B$ is orthogonal however, $B^TB=I$ and so $$\text{det}(B^TB)=\text{det}(B^T)\text{det}(B)=[\text{det}(B)]^2=\text{det}(I)=1$$ thus $\text{det}(B)= \pm 1$ and so $B$ can not be both orthogonal and skew-symmetric in odd dimensions.


A general classification: Note that any such matrix is normal, hence unitarily diagonalizable. Any orthogonal matrix is unitary, so all of it's eigenvalues are contained in the unit circle $C=\{z:|z|=1\}$ of the complex plane. Similarly any skew-symmetric has eigenvalues in $i \Bbb R$, i.e., purely imaginary. Thus the orthogonal, skew-symmetric matrices are precisely those matrices whose eigenvalues lie in $C \cap i \Bbb R = \{-i,i\}$, and which are unitarily diagonalizable.

In odd dimensions, the characteristic polynomial has at least one real root, hence no such matrices exist. In even dimensions, such matrices are precisely those whose eigenbasis is orthonormal, and whose characteristic polynomial is $(z-i)^{d_1}(z+i)^{d_2}$ where $d_1+d_2=d$. If we want the matrix to have real coefficients, then necessarily $d_1=d_2=d/2$, and thus the characteristic polynomial is $(z^2+1)^{d/2}$. This implies (by orthonormal diagonalizability), that the matrix mentioned by @Alex is the only real skew symmetric and orthogonal matrix, up to an orthogonal change of basis (i.e. conjugation in $O(d, \Bbb R)$).


I like $$ \begin{pmatrix} 0 & \mathrm{i} \\ -\mathrm{i} & 0 \end{pmatrix} \text{.} $$

(Following edit: This matrix agrees with the content of Shalop's answer and I suspect is pointing in the direction J.M. is not a mathematician was thinking in his comment.)

  • $\begingroup$ This matrix is not skew-symmetric at all! (It's symmetric instead... $A^T = A$, not $-A$.) $\endgroup$ – Ramillies Aug 28 '17 at 11:32
  • $\begingroup$ @Ramillies It does satisfy $A^*=-A$ since the Hermitian adjoint $A^*$ is defined to be the conjugate transpose for any complex matrix $A$. $\endgroup$ – Shalop Aug 28 '17 at 11:49
  • $\begingroup$ That's completely true, but the question asked for a matrix satisfying $A^T = -A$, not $A^* = -A$. $\endgroup$ – Ramillies Aug 28 '17 at 12:00
  • 2
    $\begingroup$ @Ramillies is on to me. I haven't touched a non-Hermitian matrix in so long my fingers refused to type one. Fixed. $\endgroup$ – Eric Towers Aug 28 '17 at 12:10
  • $\begingroup$ @Ramillies My answer also pertains to Hermitian adjoints and not necessarily transposes, so I guess it's wrong as well. Nevertheless, said theory is still useful in classifying the ones with real entries. $\endgroup$ – Shalop Aug 28 '17 at 12:29

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