Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 39C / a / Solution

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 $B$.