 Solution

True. All three eigenvalues are distinct in $\mathbb{C}$, so their nonzero eigenvectors are linearly independent. Here is a direct proof of that assertion. If $a_1{v}_1+a_2{v}_2+a_3{v}_3=0$, apply $(A-{\lambda }_{2}I)(A-{\lambda }_{3}I)$ to obtainDistinctness makes both scalar factors nonzero, hence $a_1=0$. Applying the analogous products gives $a_2=a_3=0$. The three eigenvectors therefore form a basis of ${\mathbb{C}}^3$. Taking them as the columns of $P$ givesThis proves that every such matrix is a diagonalizable matrix over $\mathbb{C}$, rather than merely showing it for the example.