An operator $T:V\to V$ is diagonalizable iff $V$ has a basis consisting of eigenvectors of $T$. If $T(b_j)=\lambda_jb_j$, its matrix in that basis is $\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Conversely, a diagonal matrix representation supplies those eigenvector relations.
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