Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 4 / b / ii / Solution

Take the vector subspace of skew-symmetric matrices,It is closed under vector addition and scalar multiplication, and a basis isIndeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence $\boxed{\dim K=\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. Part (a)(ii) proves that the only diagonalisable matrix in $K$ is $0_n$.