Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 7B (Vectors and Matrices)

If $A^T=-A$, then the scalar $x^{T}Ax$ equals its own transpose and henceso it vanishes.

Conversely, suppose $x^{T}Ax=0$ for every real $x$. Taking $x=e_i$ gives $A_{ii}=0$. Taking $x=e_i+e_j$ then gives $A_{ij}+A_{ji}=0$. Thus every entry satisfies $A^T=-A$, and

A real skew-symmetric matrix satisfies $A^{†}=A^T=-A$, so it is a skew-Hermitian matrix. If $Ax=\lambda x$, thenTaking the complex conjugate and using $A^{†}=-A$ shows that this scalar is the negative of its conjugate. Henceso every eigenvalue is purely imaginary or zero.

If $Ax=\lambda x$ and $Ay=\mu y$, thenSince ${\lambda }^{\ast }=-\lambda$, this becomes $(\mu -\lambda )x^{†}y=0$. Distinct eigenvalues therefore have orthogonal eigenvectors:

In odd dimension,so $\det A=0$. Since $A$ is real, its kernel contains a nonzero real vector $a$.

The plane $a^{\perp }$ is invariant under $A$, because $a\cdot Ab=-(Aa)\cdot b=0$. On this two-dimensional plane, $Ab$ is perpendicular to $b$. Put $|Ab|=\theta |b|$ with $\theta >0$. Then $A^{2}b$ is parallel to $b$, andsoThe matrix exponential fixes the kernel vector:Separating even and odd powers in its series and using $A^{2j}b=(-1{)}^j{\theta }^{2j}b$ givesThus $e^A$ acts as a rotation through angle $\theta$ on $a^{\perp }$ and fixes its axis $\mathbb{R}a$.