 Solution

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 $n=1$.