= Solution
A matrix is <diagonalizable> if some invertible $S$ makes $S^{-1}AS$ diagonal, equivalently if it has a basis of <eigenvectors>. If $A^p=0$ and $Av=\lambda v$ for $v\ne0$, then $0=A^pv=\lambda^pv$, so every <eigenvalue> is zero. There is at least one zero eigenvalue: otherwise $A$ and hence $A^p$ would be invertible. A <diagonalizable nilpotent matrix> has an all-zero diagonal form and therefore is zero itself.
For the specified matrices, direct determinant expansion gives
$$
\boxed{\det(\lambda I-B)=\lambda^3,\qquad\det(\lambda I-C)=\lambda(\lambda-6)(\lambda+6).}
$$
The <Cayley-Hamilton theorem> gives $B^3=0$. Its entry $B_{12}=-2$ is nonzero for every real $\alpha$, so \b[$B$ is not diagonalizable for any $\alpha$]. The matrix $C$ has three distinct real eigenvalues and is therefore \b[diagonalizable over $\mathbb R$]. For an explicit check, take columns $(-1,0,1)^T$, $(2,-3,1)^T$, $(2,3,1)^T$ in $S$; then $S^{-1}CS=\operatorname{diag}(-6,0,6)$.
Back to article page
