There exists an orthogonal antisymmetric real $n\times n$-matrix iff $n$ is even

Question

How do I prove that there exists an orthogonal antisymmetric $n\times n$-matrix with real coefficients iff $n$ is even?

I know that orthogonal and antisymmetric means $AA^T=I$ and $A=-A^T$, thus $A^2=-I$.

Answer 1

You can answer this by computing determinants. First, assume $A$ is orthogonal and anti-symmetric. Then we have

$$1 = \det{(I)} = \det{(A A^T)} = \det{(A)}\det{(A^T)}=(\det{A})^2.$$

Hence $\det{A} = \pm 1$ because $A$ is real. On the other hand, antisymmetry gives you $A = -A^T$, thus

$$\det{(A)} = \det{(-A^T)} = (-1)^n \det{(A^T)} = (-1)^n \det{(A)} $$

Canceling the (non-zero) determinant leaves you with $1 = (-1)^n$, which is only ture when $n$ is even.

So we see, that if a real matrix is invertible and antisymmetric, it must be of even dimension. (Orthogonality is not needed for this direction.)

Conversely, if $n= 2k$ is even, build a $n \times n$-Matrix with $k$ $2\times 2$-block-matrices of the form $$ \begin{pmatrix}0 &-1 \\1 &0\end{pmatrix} $$ around the diagonal. This matrix will be anti-symmteric and orthogonal since the columns will form an orthonormal basis of $\mathbb{R}^n$.

Answer 2

If $A^2=-I$, then the only eigenvalues of $A$ are $\lambda=\pm i$. Since $A$ is a real matrix, so is its characteristic polynomial $p_A$, and hence the multiplicity $m$ of $\lambda =i$ is equal to the multiplicity of $\lambda=-i$. Thus the degree of $p_A$ is $2m$ and so is the dimension of $A$.