Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 7B (Vectors and Matrices)

Expansion along the first column gives the characteristic polynomialThus 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 $\mathrm{span}\{(1,0,0{)}^T\}$ and has geometric multiplicity one. A diagonalisable matrix needs an eigenvector basis, so $M$ is not diagonalisable.

If $B=S^{-1}AS$ with $S$ invertible, then $tI-B=S^{-1}(tI-A)S$. Multiplicativity of the determinant givesThe roots of the characteristic polynomial, counted with their multiplicities, are the eigenvalues with their algebraic multiplicities. Thus similar matrices have the same eigenvalues with the same algebraic multiplicities.

The converse is false. The matricesboth have characteristic polynomial $(t-1{)}^2$, but every matrix similar to $I_2$ is $S^{-1}{I}_{2}S=I_2$, whereas $J\ne I_2$. Alternatively, their geometric multiplicities for $1$ differ.

The Cayley-Hamilton theorem says that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a $2\times 2$ diagonalisable matrix, write $A=SD{S}^{-1}$ with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. The characteristic polynomial is $(t-{\lambda }_1)(t-{\lambda }_2)$, including repeated roots. Evaluating it at $D$ gives zero in each diagonal position, and polynomial evaluation respects matrix similarity: ${\chi }_A(A)=S{\chi }_A(D)S^{-1}=0$. This proves the requested case.

For the final assertion, let $\lambda$ be any complex eigenvalue of $B$, with nonzero eigenvector $\mathbf{v}$. From $B^k=0$ we obtain $0=B^k\mathbf{v}={\lambda }^k\mathbf{v}$, so $\lambda =0$. The characteristic polynomial of this nilpotent linear map is therefore $t^n$. Applying the full Cayley-Hamilton theorem yields $\boxed{B^n=0}$. This argument does not assume $B$ is diagonalisable.