Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 39C (Numerical Analysis)

Because $Sw=c{e}_1$ with $c\ne 0$, the similarity transform satisfies $\hat{A}{e}_1={\lambda }_1{e}_1$. Thus its first column is $({\lambda }_1,0,\dots ,0{)}^T$ and it has block form $\left(\begin{array}{cc}{\lambda }_1& \ast \\ 0& B\end{array}\right)$. Taking the determinant gives $\det (zI-\hat{A})=(z-{\lambda }_1)\det (zI-B)$. Similarity preserves the characteristic polynomial. Hence the eigenvalues of $A$, with algebraic multiplicities, are ${\lambda }_1$ and those of $B$.

The two independent columns of $V$ remain independent after multiplication by invertible $S$. Since $R=SV$ has zeros below the second row, its leading two-by-two triangular block is invertible. Hence the image under $S$ of the invariant subspace is exactly $\mathrm{span}(e_1,e_2)$. This coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$, forcing the bottom-left block to vanish:The block determinant factors $\det (zI-\hat{A})=\det (zI-B)\det (zI-D)$. Therefore the eigenvalues of $A$ are exactly those of the indicated two-by-two and $(n-2)$-by-$(n-2)$ blocks, counted with algebraic multiplicity. No diagonalizability assumption is needed.