Nested contrast models and projections in a balanced block design

Let $H=\{a\in {\mathbb{R}}^t:{\mathbf{1}}_t^{T}a=0\}$. The coordinate-sum map ${\mathbb{R}}^t\to \mathbb{R}$ is onto, so [F2] gives $\dim H=t-1$. By [F6], $\dim \mathrm{row}(C)=\dim \mathrm{im}C=\mathrm{rank}C=q$. For every $a\in {\mathbb{R}}^t$, the vector $h=a-({\mathbf{1}}_t^{T}a\mathrm{/}t){\mathbf{1}}_t$ lies in $H$ and satisfies $Ch=Ca$, because $C{\mathbf{1}}_t=0$. Thus $\mathrm{im}(C{\mathrm{\mid }}_H)=\mathrm{im}C$ has dimension $q$, and [F2] gives $\dim (H\cap \ker C)=t-1-q$. The row space of $C$ lies in $H$ and is orthogonal to $H\cap \ker C$. The orthogonal complement of $H\cap \ker C$ within $H$ has dimension $q$: both subspaces are finite-dimensional by [F4], and [F3]–[F5] subtract dimensions from $\dim H=t-1$. Since this complement contains the $q$-dimensional row space, the equality case of [F4] identifies it with that row space.

In $V$, the constant arrays form a one-dimensional space $G$, the arrays $(a_i{)}_{ij}$ for $a\in H$ form a $(t-1)$-dimensional treatment space $T$, and the arrays $(d_j{)}_{ij}$ for $d\in {\mathbb{R}}^b$ with ${\sum }_j{d}_j=0$ form a $(b-1)$-dimensional block space $B$. The replication maps from the two zero-sum coordinate spaces to $T$ and $B$ are linear bijections: their repeated rows or columns recover the original coordinates, so they transport bases and give the stated dimensions; the block zero-sum space has dimension $b-1$ by [F2] applied to its onto coordinate-sum map. These spaces are pairwise orthogonal: the treatment-block inner product is ${\sum }_{i,j}{a}_i{d}_j=({\sum }_i{a}_i)({\sum }_j{d}_j)=0$, and either zero-sum condition also makes the relevant inner product with $G$ zero. Their sum is direct: if three mutually orthogonal vectors sum to zero, taking the inner product with each one gives its squared norm zero. Thus $W_U=G\oplus T\oplus B$, and the finite-dimensional direct-sum formula [F5] gives $\dim W_U=1+(t-1)+(b-1)=t+b-1$. The reduced space replaces $T$ by the replicated arrays from $H\cap \ker C$, whose dimension is $t-1-q$ by step 1.1; the same formula gives $\dim W_R(C)=1+(t-1-q)+(b-1)=t+b-1-q$.

For any $y$, set $r_{ij}=y_{ij}-{\bar{y}}_{i\cdot }-{\bar{y}}_{\cdot j}+{\bar{y}}_{\cdot \cdot }$. Summing $r$ over any row gives zero because ${\sum }_j{y}_{ij}=b{\bar{y}}_{i\cdot }$ and ${\sum }_j{\bar{y}}_{\cdot j}=b{\bar{y}}_{\cdot \cdot }$; summing over any column likewise gives zero. Conversely, if $v$ is orthogonal to $T$, then its row-sum vector is orthogonal to every $a\in H$, so all row sums are equal; orthogonality to $G$ makes their total zero, so every row sum is zero. The same argument with $B$ gives zero column sums. Thus $W_U^{\perp }=E$. Also $(y-r{)}_{ij}={\bar{y}}_{i\cdot }+{\bar{y}}_{\cdot j}-{\bar{y}}_{\cdot \cdot }$ belongs to $W_U$. By [F3], this is the orthogonal decomposition of $y$ relative to $W_U$, proving the formula for $RS{S}_U$; the subspace $E$ is finite-dimensional by [F4], and [F3] gives $V=W_U\oplus E$. Since the standard cell-coordinate basis makes $\dim V=tb$, [F5] gives $\dim E=tb-(t+b-1)=(t-1)(b-1)$.

Put $P_C=C^T(C{C}^T{)}^{-1}C$. The matrix $C{C}^T$ is invertible since $C$ has full row rank. Direct multiplication gives $P_C^T=P_C$, $P_C^2=P_C$, and image $P_C=\mathrm{row}(C)$. For any $x$ and $a\in {\mathbb{R}}^q$, $⟨x-P_{C}x,C^{T}a⟩=a^{T}Cx-a^{T}C{C}^T(C{C}^T{)}^{-1}Cx=0$, so [F3] identifies $P_C$ as the orthogonal projection onto $\mathrm{row}(C)$. Consequently $T$ is the orthogonal direct sum of the reduced treatment space and the treatment arrays formed from $\mathrm{row}(C)$. Therefore $W_U=W_R(C)\oplus D_C$, where $D_C=\{(a_i{)}_{ij}:a\in \mathrm{row}(C)\}$.

The projection of $y$ onto $D_C$ is the array repeating $P_C\bar{y}$ across all $b$ blocks: its residual is orthogonal to every repeated $a\in \mathrm{row}(C)$ because the inner product is $b{a}^T(\bar{y}-P_C\bar{y})=0$. Its squared norm is $b\mathrm{\parallel }{P}_C\bar{y}{\mathrm{\parallel }}^2=b(C\bar{y}{)}^T(C{C}^T{)}^{-1}(C\bar{y})$. In the orthogonal direct sum of step 3.2, $y-P_{W_R}y$ is this projection plus $y-P_{W_U}y$, and the two parts are orthogonal; expanding the squared norm gives the claimed residual-sum reduction.