Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 8A (Vectors and Matrices)

A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

If $Av=\lambda v$, then $p(A)v=p(\lambda )v$, proving one inclusion. Conversely, if $\mu$ is an eigenvalue of $p(A)$, factor $p(z)-\mu =c{\prod }_j(z-{\lambda }_j)$. If none of the ${\lambda }_j$ were eigenvalues of $A$, every $A-{\lambda }_{j}I$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu =p(\lambda )$ for some eigenvalue $\lambda$ of $A$.

Write $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$. The power series gives $e^A=S\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})S^{-1}$. Therefore

The vectors $(1,0,1{)}^T,(0,1,0{)}^T,(1,0,-1{)}^T$ are eigenvectors of $B$ with eigenvalues $2,1,0$. Applying the exponential to these eigenspaces gives