Because $S w=ce_1$ with $c\ne0$, the similarity transform satisfies $\widehat A e_1=\lambda_1e_1$. Thus its first column is $(\lambda_1,0,\ldots,0)^T$ and it has block form $\begin{pmatrix}\lambda_1&*\\0&B\end{pmatrix}$. Taking the determinant gives $\det(zI-\widehat A)=(z-\lambda_1)\det(zI-B)$. Similarity preserves the <characteristic polynomial>. Hence \b[the <eigenvalues> of $A$, with algebraic multiplicities, are $\lambda_1$ and those of $B$].

Solution

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

 Solution (source code)