Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Digital systems, linear algebra, introduction to formal proof / 6 / a

Since $B$ is upper triangular, its eigenvalues are its diagonal entries. Equivalently, its characteristic polynomial is $\lambda (\lambda -3/8)$. Hence

For eigenvalue zero take $s_1=(1,0{)}^T$. For eigenvalue $3/8$, the first row of $(B-3I/8)s_2=0$ gives $x=-2y$, so take $s_2=(-2,1{)}^T$. These are linearly independent eigenvectors, givingThe identities $B{s}_i={\lambda }_i{s}_i$ give $BS=SD$, hence the required diagonalization $S^{-1}BS=D$.