 Solution

Yes. Consider the following $n^2$ matrices: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 $E_{ii}$ andThus 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.