Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 4 / c (A basis of diagonalisable matrices)

Write $r=(r_1,\dots ,r_n{)}^T$. The linear map represented by $A$ has the formThe image of a linear map here is contained in the one-dimensional linear span of $e_i$, so its matrix rank is at most one. The rank-nullity theorem yieldsThus the solution space of $Ax=0$, namely the zero eigenspace when zero is an eigenvalue, contains at least $n-1$ linearly independent vectors. If $r=0$ its dimension is $n$; otherwise it is $n-1$. When $n=1$ the asserted lower bound is zero, even if zero is not an eigenvalue.

Since $A{e}_i=r_i{e}_i$, the vector $e_i$ is an eigenvector with nonzero eigenvalue $r_i$. The assumption $r_i\ne 0$ also makes $r\ne 0$, so the kernel has dimension $n-1$. Choose its basis $u_1,\dots ,u_{n-1}$.

The vector $e_i$ is outside this kernel. Hence $(u_1,\dots ,u_{n-1},e_i)$ is a basis of ${\mathbb{R}}^n$ consisting of eigenvectors, with corresponding eigenvalues $0,\dots ,0,r_i$. In that basis, $A$ is the diagonal matrixThis proves that $A$ is a diagonalisable matrix, including the case $n=1$.

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.