= Solution
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 $\operatorname{span}(e_1,e_2)$. This coordinate plane is invariant under $\widehat A=SAS^{-1}$, forcing the bottom-left block to vanish:
$$
\widehat A=\begin{pmatrix}B&C\\0&D\end{pmatrix}.
$$
The block determinant factors $\det(zI-\widehat A)=\det(zI-B)\det(zI-D)$. Therefore \b[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.
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