[proofplan] We verify $SAS^{-1}e_1 = \lambdae_1$ by computing $S^{-1}e_1 = w/w_1$, then applying $A$ (using the eigenvalue equation $Aw = \lambdaw$), and finally applying $S$ (using the explicit form of the elementary elimination matrix). The proof requires showing that $Sw = w_1e_1$, which follows from the structure of $S$. [/proofplan] [step:Compute $S^{-1}e_1$ from the structure of the elimination matrix] The matrix $S$ is the elementary elimination matrix with $S_{ij} = \delta_{ij}$ for $i = 1$ or $j \neq 1$, and $S_{i1} = -w_i/w_1$ for $i \geq 2$. Its inverse $S^{-1}$ has the same structure but with the signs of the subdiagonal entries in the first column reversed: $(S^{-1})_{i1} = w_i/w_1$ for $i \geq 2$, and $(S^{-1})_{ij} = \delta_{ij}$ otherwise. The first column of $S^{-1}$ is therefore: \begin{align*} S^{-1}e_1 = \begin{pmatrix} 1 \\ w_2/w_1 \\ \vdots \\ w_n/w_1 \end{pmatrix} = \frac{1}{w_1}w. \end{align*} [/step] [step:Apply $A$ using the eigenvalue equation and then $S$ to verify $SAS^{-1}e_1 = \lambdae_1$] Substituting $S^{-1}e_1 = w/w_1$: \begin{align*} SAS^{-1}e_1 = SA\left(\frac{w}{w_1}\right) = \frac{1}{w_1}S(Aw) = \frac{\lambda}{w_1}Sw, \end{align*} where the last equality uses $Aw = \lambdaw$. It remains to compute $Sw$. For the first component: \begin{align*} (Sw)_1 = \sum_{j=1}^n S_{1j} w_j = w_1, \end{align*} since the first row of $S$ is $e_1^\top$. For $i \geq 2$: \begin{align*} (Sw)_i = S_{i1}w_1 + \sum_{j=2}^n S_{ij}w_j = -\frac{w_i}{w_1} \cdot w_1 + w_i = -w_i + w_i = 0, \end{align*} since $S_{ij} = \delta_{ij}$ for $j \geq 2$ and $S_{i1} = -w_i/w_1$. Therefore $Sw = w_1e_1$, and: \begin{align*} SAS^{-1}e_1 = \frac{\lambda}{w_1} \cdot w_1e_1 = \lambdae_1. \end{align*} [guided] The strategy is the same as in the Householder deflation proof: we show that $SAS^{-1}$ maps $e_1$ to $\lambdae_1$ by chaining three operations — $S^{-1}$ recovers (a scalar multiple of) the eigenvector $w$, then $A$ multiplies it by $\lambda$, then $S$ maps the result back to a multiple of $e_1$. The computation $Sw = w_1e_1$ captures the essential purpose of the elimination matrix $S$: its subdiagonal entries in the first column are chosen precisely so that $-\frac{w_i}{w_1}w_1 + w_i = 0$ for each $i \geq 2$, zeroing out all components of $w$ below the first. This is the same mechanism as Gaussian elimination — $S$ performs the row operations that eliminate the entries below the pivot $w_1$. Combining the three factors: \begin{align*} SAS^{-1}e_1 = SA\left(\frac{w}{w_1}\right) = \frac{1}{w_1}SAw = \frac{\lambda}{w_1}Sw = \frac{\lambda}{w_1}(w_1e_1) = \lambdae_1. \end{align*} Note that this argument requires $w_1 \neq 0$ for $S$ to be well-defined. When $w_1 = 0$, one must first permute the components of $w$ to bring a nonzero entry to the first position before constructing $S$. [/guided] [/step]