 Solution

Use the full $3\times 3$ matrix $B$ in the original PDF; the transcription has dropped its middle column. For either matrix, the three-dimensional characteristic polynomial is ${\chi }_M(t)=t^3-(\mathrm{tr}M)t^2+[(\mathrm{tr}M{)}^2-\mathrm{tr}(M^2)]t/2-\det M$. Both have zero trace and zero determinant, while $\mathrm{tr}(B^2)=0$ and $\mathrm{tr}(C^2)=72$. ConsequentlyFor $B$, a useful check isfor every value of $\alpha$. Thus $B$ is a nonzero nilpotent matrix for all $\alpha$; for example its second column never vanishes. By the preceding result, $\boxed{B\text{ is not diagonalizable for any }\alpha }$.

For $C$, the three eigenvalues $-6,0,6$ are distinct and real, so their eigenvectors are linearly independent. An explicit eigenbasis isWith $P=(v_{-},v_0,v_{+})$, $\det P=18\ne 0$ and $P^{-1}CP=\mathrm{diag}(-6,0,6)$. Hence $\boxed{C\text{ is diagonalizable over }\mathbb{R}\text{ and }\mathbb{C}}$.

A repeated root of the characteristic polynomial alone would not settle diagonalizability; for $B$ it is the nonzero nilpotence that completes the argument.