= Solution
{c}
Yes. Consider the following $n^2$ <matrices>:
$$
\boxed{\mathcal B=\{E_{ii}:1\le i\le n\}
\ \cup\ \{E_{ii}+E_{ij}:1\le i,j\le n,\ i\ne j\}.}
$$
Each has only row $i$ nonzero and has $i$th diagonal entry $1$, so part (c)(ii) proves that it is a <diagonalisable matrix>. The collection is a <spanning set>, because it contains every $E_{ii}$ and
$$
E_{ij}=(E_{ii}+E_{ij})-E_{ii}\qquad(i\ne j).
$$
Thus it spans the <matrix unit> <basis>. A <spanning set> of $n^2$ <vectors> in an $n^2$-dimensional <vector space> is a <basis>. Alternatively, in a vanishing <linear combination> of $\mathcal B$, each off-diagonal entry first forces its own coefficient to be zero, and then each diagonal entry forces the remaining coefficient to be zero.
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