[proofplan] We derive the row formula for the deflated matrix $B$ from the structure of $SAS^{-1}$. The key observation is that $S^{-1}$ modifies only the first column (leaving columns $2, \ldots, n$ as standard basis vectors), so the last $n-1$ columns of $SAS^{-1}$ equal those of $SA$. The $i$-th row of $SA$ for $i \geq 2$ is $A_i - (w_i/w_1)A_1$ by the elimination structure of $S$. Relabelling and dropping the first component yields the formula for $B$. [/proofplan] [step:Identify the last $n-1$ columns of $SAS^{-1}$ with those of $SA$] The inverse elimination matrix $S^{-1}$ differs from the identity only in its first column: $(S^{-1})_{i1} = w_i/w_1$ for $i \geq 2$, and $(S^{-1})_{ij} = \delta_{ij}$ for $j \geq 2$. Therefore, for $j \geq 2$, the $j$-th column of $S^{-1}$ is $e_j$, and: \begin{align*} (SAS^{-1})e_j = SAe_j = S \cdot (\text{$j$-th column of $A$}). \end{align*} The last $n-1$ columns of $SAS^{-1}$ are therefore the last $n-1$ columns of $SA$. [/step] [step:Compute the rows of $SA$ for $i \geq 2$ and extract the formula for $B$] The matrix $S$ acts on a vector $v$ by: $(Sv)_1 = v_1$ and $(Sv)_i = v_i - (w_i/w_1)v_1$ for $i \geq 2$. Equivalently, row $i$ of $SA$ (for $i \geq 2$) is: \begin{align*} (SA)_i = A_i - \frac{w_i}{w_1}A_1, \end{align*} where $A_j$ denotes the $j$-th row of $A$. Since $SAS^{-1}e_1 = \lambdae_1$ (by the [Elementary Deflation Produces Correct Block Form](/theorems/1416) result), the matrix $SAS^{-1}$ has the form: \begin{align*} SAS^{-1} = \begin{pmatrix} \lambda & r^\top \\ \mathbf{0} & B \end{pmatrix}, \end{align*} where $B \in \mathbb{R}^{(n-1) \times (n-1)}$. The $i$-th row of $B$ (for $i = 1, \ldots, n-1$) corresponds to row $i+1$ of $SAS^{-1}$, restricted to columns $2, \ldots, n$. From the previous step, these columns agree with those of $SA$, so row $i$ of $B$ is: \begin{align*} B_i = (SA)_{i+1}\big|_{\text{columns } 2, \ldots, n} = \left(A_{i+1} - \frac{w_{i+1}}{w_1}A_1\right)\bigg|_{\text{drop first component}}. \end{align*} [guided] Why does the formula involve dropping the first component? The full row $(SA)_{i+1}$ has $n$ entries and corresponds to row $i+1$ of the $n \times n$ matrix $SAS^{-1}$. But the $(i+1, 1)$ entry of $SAS^{-1}$ is $0$ (from the block structure established by the [Elementary Deflation Produces Correct Block Form](/theorems/1416) result), so the first component of $(SA)_{i+1}$ belongs to the first column of $SAS^{-1}$, not to $B$. The remaining $n-1$ entries form row $i$ of $B$. Explicitly, for each $i = 1, \ldots, n-1$: \begin{align*} B_i = A_{i+1} - \frac{w_{i+1}}{w_1}A_1, \end{align*} where both $A_{i+1}$ and $A_1$ are understood as row vectors in $\mathbb{R}^n$, and we then discard the first component of the resulting vector to obtain a row in $\mathbb{R}^{n-1}$. This formula has a direct computational interpretation: to form row $i$ of the deflated matrix $B$, take row $i+1$ of $A$ and subtract $\frac{w_{i+1}}{w_1}$ times row $1$ of $A$. This is exactly the row operation that Gaussian elimination performs to zero out the $(i+1, 1)$ entry, with $w_1$ acting as the pivot. [/guided] [/step]