[proofplan] Decompose each centered observation $X_{ij}-\bar{X}$ into its within-group part $X_{ij}-\bar{X}_i$ and its group-mean part $\bar{X}_i-\bar{X}$. Expanding the resulting outer product gives within terms, between terms, and two mixed terms. The within terms sum to $W$, the between terms sum to $B$, and the mixed terms vanish because the deviations from each group mean sum to zero within every group. [/proofplan] [step:Decompose each centered observation into within-group and between-group parts] For each $i \in \{1,\dots,g\}$ and $j \in \{1,\dots,n_i\}$, define vectors $a_{ij}, b_i \in \mathbb{R}^p$ by \begin{align*} a_{ij} := X_{ij}-\bar{X}_i, \qquad b_i := \bar{X}_i-\bar{X}. \end{align*} Then \begin{align*} X_{ij}-\bar{X} = (X_{ij}-\bar{X}_i)+(\bar{X}_i-\bar{X}) = a_{ij}+b_i. \end{align*} Therefore the total SSP matrix satisfies \begin{align*} T &= \sum_{i=1}^g\sum_{j=1}^{n_i} (a_{ij}+b_i)(a_{ij}+b_i)^\top. \end{align*} [guided] The purpose of the decomposition is to separate two sources of variation. The vector $a_{ij}:=X_{ij}-\bar{X}_i$ measures how far observation $X_{ij}$ is from its own group mean, while $b_i:=\bar{X}_i-\bar{X}$ measures how far the group mean is from the grand mean. Both vectors lie in $\mathbb{R}^p$, so their outer products are $p \times p$ matrices. For every $i \in \{1,\dots,g\}$ and $j \in \{1,\dots,n_i\}$, \begin{align*} X_{ij}-\bar{X} &= X_{ij}-\bar{X}_i+\bar{X}_i-\bar{X}\\ &= a_{ij}+b_i. \end{align*} Substituting this identity into the definition of $T$ gives \begin{align*} T &= \sum_{i=1}^g\sum_{j=1}^{n_i} (X_{ij}-\bar{X})(X_{ij}-\bar{X})^\top\\ &= \sum_{i=1}^g\sum_{j=1}^{n_i} (a_{ij}+b_i)(a_{ij}+b_i)^\top. \end{align*} This converts the theorem into a purely algebraic identity about expanding outer products and summing the resulting terms. [/guided] [/step] [step:Expand the outer products and identify the within-group contribution] For vectors $u,v \in \mathbb{R}^p$, bilinearity of matrix multiplication gives \begin{align*} (u+v)(u+v)^\top = uu^\top + uv^\top + vu^\top + vv^\top. \end{align*} Applying this identity with $u=a_{ij}$ and $v=b_i$ yields \begin{align*} T &= \sum_{i=1}^g\sum_{j=1}^{n_i} a_{ij}a_{ij}^\top + \sum_{i=1}^g\sum_{j=1}^{n_i} a_{ij}b_i^\top + \sum_{i=1}^g\sum_{j=1}^{n_i} b_i a_{ij}^\top + \sum_{i=1}^g\sum_{j=1}^{n_i} b_i b_i^\top. \end{align*} By the definition of $a_{ij}$, \begin{align*} \sum_{i=1}^g\sum_{j=1}^{n_i} a_{ij}a_{ij}^\top &= \sum_{i=1}^g\sum_{j=1}^{n_i} (X_{ij}-\bar{X}_i)(X_{ij}-\bar{X}_i)^\top\\ &= W. \end{align*} [/step] [step:Show that the mixed terms vanish within each group] For each fixed $i \in \{1,\dots,g\}$, the definition of $\bar{X}_i$ gives \begin{align*} \sum_{j=1}^{n_i} a_{ij} &= \sum_{j=1}^{n_i}(X_{ij}-\bar{X}_i)\\ &= \sum_{j=1}^{n_i}X_{ij}-n_i\bar{X}_i\\ &= \sum_{j=1}^{n_i}X_{ij}-n_i\left(\frac{1}{n_i}\sum_{j=1}^{n_i}X_{ij}\right)\\ &= 0. \end{align*} Since $b_i$ does not depend on $j$, we have \begin{align*} \sum_{j=1}^{n_i} a_{ij}b_i^\top &= \left(\sum_{j=1}^{n_i} a_{ij}\right)b_i^\top = 0b_i^\top = 0,\\ \sum_{j=1}^{n_i} b_i a_{ij}^\top &= b_i\left(\sum_{j=1}^{n_i} a_{ij}\right)^\top = b_i0^\top = 0. \end{align*} Summing these identities over $i$ gives \begin{align*} \sum_{i=1}^g\sum_{j=1}^{n_i} a_{ij}b_i^\top = \sum_{i=1}^g\sum_{j=1}^{n_i} b_i a_{ij}^\top = 0. \end{align*} [guided] The mixed terms vanish because deviations from a sample mean always sum to zero inside the group. Fix a group index $i \in \{1,\dots,g\}$. By definition, \begin{align*} \bar{X}_i = \frac{1}{n_i}\sum_{j=1}^{n_i}X_{ij}. \end{align*} Therefore \begin{align*} \sum_{j=1}^{n_i} a_{ij} &= \sum_{j=1}^{n_i}(X_{ij}-\bar{X}_i)\\ &= \sum_{j=1}^{n_i}X_{ij}-\sum_{j=1}^{n_i}\bar{X}_i\\ &= \sum_{j=1}^{n_i}X_{ij}-n_i\bar{X}_i\\ &= \sum_{j=1}^{n_i}X_{ij}-n_i\left(\frac{1}{n_i}\sum_{j=1}^{n_i}X_{ij}\right)\\ &= 0. \end{align*} The vector $b_i=\bar{X}_i-\bar{X}$ depends only on the group index $i$, not on the observation index $j$. Hence it may be factored out of the sum over $j$: \begin{align*} \sum_{j=1}^{n_i} a_{ij}b_i^\top &= \left(\sum_{j=1}^{n_i}a_{ij}\right)b_i^\top = 0b_i^\top = 0,\\ \sum_{j=1}^{n_i} b_i a_{ij}^\top &= b_i\left(\sum_{j=1}^{n_i}a_{ij}\right)^\top = b_i0^\top = 0. \end{align*} Since this holds for every group $i$, summing over all groups gives \begin{align*} \sum_{i=1}^g\sum_{j=1}^{n_i} a_{ij}b_i^\top = \sum_{i=1}^g\sum_{j=1}^{n_i} b_i a_{ij}^\top = 0. \end{align*} [/guided] [/step] [step:Identify the between-group contribution and conclude the decomposition] Because $b_i$ is independent of $j$ for fixed $i$, \begin{align*} \sum_{i=1}^g\sum_{j=1}^{n_i} b_i b_i^\top &= \sum_{i=1}^g n_i b_i b_i^\top\\ &= \sum_{i=1}^g n_i(\bar{X}_i-\bar{X})(\bar{X}_i-\bar{X})^\top\\ &= B. \end{align*} Combining the expansion of $T$, the identification of the within-group term with $W$, the vanishing of the mixed terms, and the identification of the between-group term with $B$, we obtain \begin{align*} T = W + B. \end{align*} Since matrix addition is commutative, this is equivalently \begin{align*} T = B + W. \end{align*} This proves the one-way MANOVA SSP decomposition. [/step]