 Solution

Over the complex numbers, the characteristic polynomial splits into three distinct roots of a polynomial, $1,i,-i$. Direct multiplication gives the eigenvectorsThey are linearly independent: the last two have two-dimensional determinant $2i\ne 0$ in the first two coordinates, and the first has nonzero third coordinate. ThusThis is a basis of eigenvectors of ${\mathbb{C}}^3$, proving that $T^{\prime }$ has a diagonalizable matrix.