Solution
= Solution
{c}
Take the <vector subspace> of <skew-symmetric matrices>,
$$
K=\{A\in M_n(\mathbb R):A^T=-A\}.
$$
It is closed under <vector addition> and <scalar multiplication>, and a <basis> is
$$
\{E_{ij}-E_{ji}:1\le i<j\le n\}.
$$
Indeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence $\boxed{\dim K=\binom n2}$. Part (a)(ii) proves that the only <diagonalisable matrix> in $K$ is $0_n$.