A Gaussian contrast is independent of its orthogonal residual

Put $P=P_{\eta }$. Since $\eta \mathrm{\ne }0$, $P^T=P$ and $P^2=P$. For every $y$, $Py=\eta ({\eta }^{T}y)\mathrm{/}\mathrm{\parallel }\eta {\mathrm{\parallel }}^2$ belongs to $\mathrm{span}\{\eta \}$, and ${\eta }^T(y-Py)=0$, so $y-Py\in (\mathrm{span}\{\eta \}{)}^{\perp }$. Thus $P$ is the rank-one orthogonal projection onto $\mathrm{span}\{\eta \}$ in [F4]. The conclusion concerns only the joint law of $T,Z$, so by [F1] it suffices to use the realization $Y=\mu +\sigma W$ with $W\sim N_n(0,I_n)$. Apply [F2] to $P$. This use of AC is exactly [A1].

Because $P\eta =\eta$, $T={\eta }^T\mu +\sigma {\eta }^{T}PW$, while $Z=(I_n-P)\mu +\sigma (I_n-P)W$. Thus $T$ and $Z$ are measurable functions of the independent vectors $PW$ and $(I_n-P)W$; [F3] makes them independent.

By the linear-projection clause of [F1], $T={\eta }^{T}Y$ is normal with mean ${\eta }^T\mu$ and variance ${\eta }^T({\sigma }^2{I}_n)\eta ={\sigma }^2\mathrm{\parallel }\eta {\mathrm{\parallel }}^2$. AC is inherited from [A1] through the normal-law supplier [F1].