 Solution

The characteristic polynomial has only the eigenvalue $\mu$. If $A$ is diagonalizable, there is an invertible matrix $P$ and a diagonal matrix $D$ with $A=PD{P}^{-1}$. Every diagonal entry of $D$ is an eigenvalue, hence equals $\mu$. Thus $D=\mu I$ and $A=P(\mu I)P^{-1}=\mu I$. Conversely, $\mu I$ is already diagonal in every basis. HenceThis argument uses the existence of an eigenvector basis rather than merely the repeated root of the polynomial; repeated eigenvalues alone do not imply diagonalizability.