Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 6C (Vectors and Matrices)

The characteristic polynomial isFor $\lambda =2$, the eigenspace is spanned only by $(1,0,0{)}^T$, so its geometric multiplicity is one although its algebraic multiplicity is two. Hence $M$ is not diagonalisable.

If $B=S^{-1}AS$, thenso the characteristic polynomials, eigenvalues, and algebraic multiplicities agree. The converse is false: $I_2$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ have the same characteristic polynomial, but the first is diagonalisable and the second is not, so they cannot be similar.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; applying $p(t)=(t-{\lambda }_1)(t-{\lambda }_2)$ to the diagonal matrix gives zero and hence $p(A)=0$.

If $B^k=0$, every eigenvalue $\lambda$ satisfies ${\lambda }^k=0$, so all eigenvalues are zero and the characteristic polynomial is $t^n$. Cayley--Hamilton then gives $B^n=0$.