Solution
= Solution
Expansion along the first column gives the <characteristic polynomial>
$$
\chi_M(t)=\det(tI-M)=(t-2)[(t-1)(t-4)+2]=(t-2)^2(t-3).
$$
Thus the <eigenvalue> $2$ has <algebraic multiplicity> two. Solving $(M-2I)(x,y,z)^T=0$ gives $y=0$ and then $z=0$, leaving $x$ arbitrary. Its <eigenspace> is $\operatorname{span}\{(1,0,0)^T\}$ and has <geometric multiplicity> one. A <diagonalisable matrix> needs an <eigenvector> <basis>, so \b[$M$ is not diagonalisable].