Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 4 / a (Symmetric and skew-symmetric matrices)

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

Let $A$ be a skew-symmetric matrix. For every real vector $x$, the real scalar $x^{T}Ax$ equals its own transpose, butHence $x^{T}Ax=0$. If $Av=\lambda v$ is a real eigenvector equation with $v\ne 0$, it follows thatso $\lambda =0$. If $A$ is a diagonalisable matrix over $\mathbb{R}$, it has a real basis of eigenvectors, all with eigenvalue zero. Thus $A$ vanishes on a basis and $\boxed{A=0_n}$.

The field matters: a nonzero real skew-symmetric matrix can be a diagonalisable matrix over the complex numbers, for example $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ with eigenvalues $i,-i$.