Since $S\mathbf w=c\mathbf e^{(1)}$ and $A\mathbf w=\lambda_1\mathbf w$,
$$
\widehat A(c\mathbf e^{(1)})
=SAS^{-1}S\mathbf w
=SA\mathbf w
=\lambda_1c\mathbf e^{(1)}.
$$
Thus the first column of $\widehat A$ is $(\lambda_1,0,\ldots,0)^T$, and $\widehat A$ is a <block upper triangular matrix>:
$$
\widehat A=
\begin{pmatrix}
\lambda_1&*\\
0&B
\end{pmatrix},
$$
where $B$ is its bottom-right $(n-1)\times(n-1)$ submatrix. Therefore
$$
\det(zI-\widehat A)=(z-\lambda_1)\det(zI-B).
$$
A <similarity transformation> preserves the <characteristic polynomial>, so
$$
\boxed{\operatorname{spec}(A)=\{\lambda_1\}\mathbin\cup\operatorname{spec}(B),}
$$
with <algebraic multiplicities> included.

To construct $S$, use a <Householder transformation>. With $\alpha=\|\mathbf w\|_2$, choose the sign of $\alpha$ to avoid cancellation and set
$$
\mathbf v=\mathbf w-\alpha\mathbf e^{(1)},
\qquad
S=I-2\frac{\mathbf v\mathbf v^T}{\mathbf v^T\mathbf v}.
$$
Then $S$ is an <orthogonal matrix> and maps $\mathbf w$ to $\alpha\mathbf e^{(1)}$ up to the chosen sign. If $\mathbf w$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of <orthogonal coordinate reduction of a subspace>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 40E / a / Solution

For a full-column-rank $n\times k$ matrix $V$, extend the thin <QR decomposition> $V=Q_1R_1$ to an <orthogonal matrix> $Q=(Q_1\ Q_2)$. Then
$$
Q^TV=\begin{pmatrix}R_1\\0\end{pmatrix},
$$
so $Q^T$ maps the column space of $V$ onto the first $k$ coordinate directions. Successive <Householder transformations> construct the same reduction without first forming a full orthogonal basis.

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 Incoming links: Orthogonal coordinate reduction of a subspace