[proofplan]
We compute the difference $\hat{\ell}_j(x)-\hat{\ell}_k(x)$ directly from the definition of the sample LDA scores. The terms involving $x$ combine into one linear functional with normal vector $S_{\mathrm{pooled}}^{-1}(\hat{\mu}_j-\hat{\mu}_k)$, while the remaining terms are constants depending only on the fitted parameters. Setting the difference equal to zero gives the displayed affine equation, and invertibility of $S_{\mathrm{pooled}}$ together with $\hat{\mu}_j \neq \hat{\mu}_k$ ensures that the normal vector is nonzero.
[/proofplan]
[step:Subtract the two sample LDA scores and isolate the linear term]
Fix distinct indices $j,k \in \{1,\dots,G\}$ with $\hat{\mu}_j \neq \hat{\mu}_k$. For an arbitrary point $x \in \mathbb{R}^p$, the definitions of $\hat{\ell}_j$ and $\hat{\ell}_k$ give
\begin{align*}
\hat{\ell}_j(x)-\hat{\ell}_k(x)
&=
\left(x^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_j
-\frac{1}{2}\hat{\mu}_j^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_j
+\log \hat{\pi}_j\right) \\
&\quad -
\left(x^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_k
-\frac{1}{2}\hat{\mu}_k^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_k
+\log \hat{\pi}_k\right) \\
&=
x^\top S_{\mathrm{pooled}}^{-1}(\hat{\mu}_j-\hat{\mu}_k)
-\frac{1}{2}\left(\hat{\mu}_j^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_j
-\hat{\mu}_k^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_k\right)
+\log\frac{\hat{\pi}_j}{\hat{\pi}_k}.
\end{align*}
The logarithm is defined because $\hat{\pi}_j>0$ and $\hat{\pi}_k>0$.
[/step]
[step:Set the score difference equal to zero]
By definition,
\begin{align*}
x \in H_{jk}
\iff \hat{\ell}_j(x)=\hat{\ell}_k(x)
\iff \hat{\ell}_j(x)-\hat{\ell}_k(x)=0.
\end{align*}
Using the expression from the previous step, this condition is equivalent to
\begin{align*}
x^\top S_{\mathrm{pooled}}^{-1}(\hat{\mu}_j-\hat{\mu}_k)
-\frac{1}{2}\left(\hat{\mu}_j^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_j
-\hat{\mu}_k^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_k\right)
+\log\frac{\hat{\pi}_j}{\hat{\pi}_k}
=0.
\end{align*}
Rearranging the constant terms to the right-hand side gives
\begin{align*}
x^\top S_{\mathrm{pooled}}^{-1}(\hat{\mu}_j-\hat{\mu}_k)
=
\frac{1}{2}\left(\hat{\mu}_j^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_j
-\hat{\mu}_k^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_k\right)
-\log\frac{\hat{\pi}_j}{\hat{\pi}_k}.
\end{align*}
Thus $H_{jk}$ is exactly the set of points satisfying the displayed affine equation.
[/step]
[step:Verify that the affine equation defines a hyperplane]
Define the vector
\begin{align*}
a_{jk}:=S_{\mathrm{pooled}}^{-1}(\hat{\mu}_j-\hat{\mu}_k)\in\mathbb{R}^p
\end{align*}
and the scalar
\begin{align*}
b_{jk}:=
\frac{1}{2}\left(\hat{\mu}_j^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_j
-\hat{\mu}_k^\top S_{\mathrm{pooled}}^{-1}\hat{\mu}_k\right)
-\log\frac{\hat{\pi}_j}{\hat{\pi}_k}\in\mathbb{R}.
\end{align*}
Since $S_{\mathrm{pooled}}$ is invertible, the [linear map](/page/Linear%20Map) $S_{\mathrm{pooled}}^{-1}:\mathbb{R}^p\to\mathbb{R}^p$ is injective. Because $\hat{\mu}_j-\hat{\mu}_k \neq 0$, injectivity gives
\begin{align*}
a_{jk}=S_{\mathrm{pooled}}^{-1}(\hat{\mu}_j-\hat{\mu}_k)\neq 0.
\end{align*}
Therefore
\begin{align*}
H_{jk}=\{x\in\mathbb{R}^p:x^\top a_{jk}=b_{jk}\}
\end{align*}
is an affine hyperplane in $\mathbb{R}^p$ with nonzero normal vector $a_{jk}$. This is precisely the claimed pairwise linear decision boundary.
[/step]
