Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Digital systems, linear algebra and introduction to formal proof / 6

Let $E_{ij}$ denote a matrix unit, with its only nonzero entry equal to one in row $i$, column $j$. A basis for the vector space of real skew-symmetric matrices isEach upper-triangular entry is free, its reflected lower-triangular entry is its negative, and every diagonal entry is zero. This is the basis of the space of real skew-symmetric matrices.

Using linearity and involutivity of the matrix transpose,Therefore $B-B^{\top }$ is a skew-symmetric matrix.

Split $B$ into its symmetric part and skew-symmetric part:Taking the transpose gives $S^{\top }=S$ and $K^{\top }=-K$. Thus $S$ is a symmetric matrix and $K$ a skew-symmetric matrix. These formulas also show uniqueness, since a real matrix that is both symmetric and skew-symmetric is zero.

If $A^{\top }=-A$, then the dot product satisfiesConversely, the assumed dot product identity gives $x^{\top }(A^{\top }+A)y=0$ for every $x,y$. Take $x=e_i$ and $y=e_j$ from the standard basis: every entry of $A^{\top }+A$ is zero. Hence $A^{\top }=-A$, proving the equivalence.

If $A$ is a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its own transpose, soand it is zero over the real numbers.

Conversely, suppose the quadratic form $x^{\top }Ax$ vanishes for every real vector. Substituting $x=e_i$ gives $a_{ii}=0$. Substituting $x=e_i+e_j$ gives $a_{ij}+a_{ji}=0$ for $i\ne j$. Therefore $A^{\top }=-A$. This proves the quadratic-form criterion for a skew-symmetric matrix:

For a real eigenvalue $\lambda$, choose a nonzero real eigenvector $v$ with $Av=\lambda v$. By (e),Since $v^{\top }v=\Vert v{\Vert }^2>0$, $\lambda =0$. If eigenvectors were initially allowed to be complex, the real matrix $A-\lambda I$ has a nonzero real null space whenever it has a nonzero complex null space, so this choice of $v$ remains justified.

The determinant is invariant under transpose, while multiplying all $n$ rows by $-1$ multiplies it by $(-1{)}^n$. Thus, for odd integer $n$,Over the real numbers, this forces $\det A=0$, so $A$ is a singular matrix. This is the odd-dimensional skew-symmetric singularity theorem.

If $(I+A)x=0$, take its dot product with $x$ and apply (e):Hence $x=0$. A square matrix with zero null space is an invertible matrix by the rank-nullity theorem. Therefore $I+A$ is nonsingular. This proves invertibility of identity plus a real skew-symmetric matrix.