[proofplan]
We verify the block structure of $HAH^\top$ in two stages. First, we show that the first column of $HAH^\top$ is $\lambdae_1$ by using $H^\tope_1 = w/c$ (from the [Householder Reflection Maps Eigenvector to First Basis Vector](/theorems/1414) result) and the eigenvalue equation $Aw = \lambdaw$. Second, we use the symmetry of $HAH^\top$ (inherited from $A$ via orthogonal conjugation) to conclude that the first row also has zeros outside the $(1,1)$ entry, giving the stated block form.
[/proofplan]
[step:Establish that $H$ is orthogonal and compute $H^\tope_1$]
The Householder matrix $H = I - 2uu^\top/\|u\|^2$ satisfies:
\begin{align*}
H^\top = \left(I - \frac{2uu^\top}{\|u\|^2}\right)^\top = I - \frac{2uu^\top}{\|u\|^2} = H,
\end{align*}
so $H$ is symmetric. Moreover:
\begin{align*}
H^2 = \left(I - \frac{2uu^\top}{\|u\|^2}\right)\left(I - \frac{2uu^\top}{\|u\|^2}\right) = I - \frac{4uu^\top}{\|u\|^2} + \frac{4u(u^\topu)u^\top}{\|u\|^4} = I.
\end{align*}
Therefore $H = H^\top = H^{-1}$, confirming that $H$ is orthogonal and involutory.
By the [Householder Reflection Maps Eigenvector to First Basis Vector](/theorems/1414) result, $Hw = ce_1$ for some nonzero $c$. Applying $H^{-1} = H$ to both sides:
\begin{align*}
w = cHe_1 = cH^\tope_1,
\end{align*}
so $H^\tope_1 = w/c$.
[/step]
[step:Compute the first column of $HAH^\top$ using the eigenvalue equation]
The first column of $HAH^\top$ is $HAH^\tope_1$. Substituting $H^\tope_1 = w/c$:
\begin{align*}
HAH^\tope_1 = HA\left(\frac{w}{c}\right) = \frac{1}{c}H(Aw) = \frac{1}{c}H(\lambdaw) = \frac{\lambda}{c}Hw = \frac{\lambda}{c}(ce_1) = \lambdae_1.
\end{align*}
This means the first column of $HAH^\top$ is $\lambdae_1$: the $(1,1)$ entry is $\lambda$ and all entries $(HAH^\top)_{i1} = 0$ for $i \geq 2$.
[/step]
[step:Use symmetry to deduce the full block structure]
Since $A$ is symmetric and $H$ is orthogonal:
\begin{align*}
(HAH^\top)^\top = (H^\top)^\top A^\top H^\top = HAH^\top,
\end{align*}
so $HAH^\top$ is symmetric. From the previous step, $(HAH^\top)_{i1} = 0$ for all $i \geq 2$. By symmetry, $(HAH^\top)_{1j} = (HAH^\top)_{j1} = 0$ for all $j \geq 2$. Therefore:
\begin{align*}
HAH^\top = \begin{pmatrix} \lambda & \mathbf{0}^\top \\ \mathbf{0} & \tilde{A} \end{pmatrix},
\end{align*}
where $\tilde{A} \in \mathbb{R}^{(n-1) \times (n-1)}$ is the lower-right block. Since $HAH^\top$ is symmetric, $\tilde{A}$ is also symmetric.
[guided]
The block structure emerges from two independent facts working together.
First, the eigenvalue equation and the Householder property give us the first column: $HAH^\tope_1 = \lambdae_1$. This tells us the first column has $\lambda$ in position 1 and zeros everywhere below. In matrix terms, this means:
\begin{align*}
HAH^\top = \begin{pmatrix} \lambda & r^\top \\ \mathbf{0} & \tilde{A} \end{pmatrix}
\end{align*}
for some row vector $r^\top \in \mathbb{R}^{1 \times (n-1)}$ and some matrix $\tilde{A} \in \mathbb{R}^{(n-1) \times (n-1)}$. At this point, we do not yet know that $r = \mathbf{0}$.
Second, symmetry forces $r = \mathbf{0}$. Since $A$ is symmetric ($A^\top = A$) and $H$ is orthogonal ($H^\top = H^{-1}$), the product $HAH^\top$ is symmetric: $(HAH^\top)^\top = HA^\top H^\top = HAH^\top$. In a symmetric matrix, the first row equals the transpose of the first column. Since the first column is $(\lambda, 0, \ldots, 0)^\top$, the first row must be $(\lambda, 0, \ldots, 0)$, so $r = \mathbf{0}$.
The symmetry of $\tilde{A}$ follows from the symmetry of the full matrix: if $B$ is symmetric and has block form $\begin{pmatrix} \lambda & \mathbf{0}^\top \\ \mathbf{0} & \tilde{A} \end{pmatrix}$, then $\tilde{A}^\top = \tilde{A}$.
[/guided]
[/step]
