Diagonalizable upper triangular matrices
Solution 1:
Every upper triangular matrix with distinct elements on the diagonal is diagonalizable, because
$$\det(A-\lambda I)=\prod_{i=1}^n (a_{ii}-\lambda)$$
with $a_{ii}\neq a_{jj}$ for $i\neq j$, so every eigenvalue has multiplicity $1$.
The converse is not true. Take $A=I$. Then $A$ is diagonalized, but not with distinct values on the diagonal.