= Solution
{c}
Let $S$ be the <vector subspace> of <symmetric matrices>. A <basis> consists of $E_{ii}$ for $1\le i\le n$ and $E_{ij}+E_{ji}$ for $i<j$, so
$$
\dim S=n+\binom n2=\frac{n(n+1)}2.
$$
The <dimension formula for a sum of subspaces> now gives
$$
\begin{aligned}
\dim(V\cap S)
&=\dim V+\dim S-\dim(V+S)\\
&\ge\dim V+\frac{n(n+1)}2-n^2\\
&=\dim V-\binom n2>0.
\end{aligned}
$$
Choose $0\ne A\in V\cap S$. It belongs to $V$ and is a <symmetric matrix>, so part (a)(i) makes it a <diagonalisable matrix>. The preceding example shows that the strict threshold is sharp.
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