Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 3 / 8B (Algebra and Geometry)

Using the monic convention for the characteristic polynomial, expand the determinant along the last column:For the eigenvalue $1$, the equations $(A-I)v=0$ give $v_2=v_1$ and $v_3=0$. Thus every corresponding nonzero eigenvector is a nonzero multiple of $(1,1,0{)}^T$. For the eigenvalue $2$, the first two equations give $v_2=2v_1$ and $v_1=0$, so every corresponding nonzero eigenvector is a nonzero multiple of $(0,0,1{)}^T$. ThereforeThese eigenspaces together have dimension $2$, so they cannot provide an eigenvector basis of ${\mathbb{R}}^3$. The matrix is not diagonalizable; the eigenvalue $1$ has algebraic multiplicity $2$ but geometric multiplicity $1$.

The characteristic polynomial has only the eigenvalue $\mu$. If $A$ is diagonalizable, there is an invertible matrix $P$ and a diagonal matrix $D$ with $A=PD{P}^{-1}$. Every diagonal entry of $D$ is an eigenvalue, hence equals $\mu$. Thus $D=\mu I$ and $A=P(\mu I)P^{-1}=\mu I$. Conversely, $\mu I$ is already diagonal in every basis. HenceThis argument uses the existence of an eigenvector basis rather than merely the repeated root of the polynomial; repeated eigenvalues alone do not imply diagonalizability.