Cellular decomposition matrices are unitriangular for the cell order

Annihilation: if $\lambda ,\nu \in \mathrm{\Lambda }$ with $\lambda ⋭\nu$, then $A^{⊵\nu }\cdot C^{\lambda }=0$. Indeed $A^{⊵\nu }$ and $A^{⊵\lambda }$ are two-sided ideals by [F3], and their intersection is spanned by the $c^{\gamma }$ with $\gamma ⊵\nu$ and $\gamma ⊵\lambda$; no such $\gamma$ equals $\lambda$, since $\gamma =\lambda$ would give $\lambda ⊵\nu$, so the intersection lies in $A^{⊳\lambda }$. Hence $A^{⊵\nu }{A}^{⊵\lambda }\subseteq A^{⊵\nu }\cap A^{⊵\lambda }\subseteq A^{⊳\lambda }$ and, since $c_{tt}^{\lambda }\in A^{⊵\lambda }$, every product $c_{tt}^{\lambda }b$ with $b\in A^{⊵\nu }$ lies in $A^{⊳\lambda }$; the right action of $b$ on $C^{\lambda }$ is read off from $c_{tt}^{\lambda }b\equiv {\sum }_v{r}_v{c}_{tv}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{⊳\lambda })$ with unique coefficients, so $b$ acts as the zero operator on $C^{\lambda }$.

Nontrivial action: if $D^{\mu }\mathrm{\ne }0$ then some element of $A^{⊵\mu }$ acts nontrivially on $D^{\mu }$. Since $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }\mathrm{\ne }{C}^{\mu }$ by [F1], there is $t\in T(\mu )$ with $c_t^{\mu }\mathrm{\notin }\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$, and by [F2] there is $u\in T(\mu )$ with $⟨c_t^{\mu },c_u^{\mu }{⟩}_{\mu }\mathrm{\ne }0$; then the element $c_{ut}^{\mu }\in A^{⊵\mu }$ satisfies $c_t^{\mu }\cdot c_{ut}^{\mu }=⟨c_t^{\mu },c_u^{\mu }{⟩}_{\mu }{c}_t^{\mu }$ by [F2], so on $D^{\mu }$ it sends the nonzero class of $c_t^{\mu }$ to its nonzero multiple $⟨c_t^{\mu },c_u^{\mu }{⟩}_{\mu }[c_t^{\mu }]\mathrm{\ne }0$.

Off-diagonal vanishing: suppose $d_{\lambda \mu }\mathrm{\ne }0$ with $D^{\mu }\mathrm{\ne }0$. If $\lambda ⋭\mu$, then step 1.1 gives $A^{⊵\mu }\cdot C^{\lambda }=0$; an element of $A$ acting as zero on a module acts as zero on each of its submodules and quotients, hence on every subquotient of $C^{\lambda }$, so $A^{⊵\mu }$ annihilates the composition factor $D^{\mu }$ of $C^{\lambda }$. This contradicts step 1.2. Hence $\lambda ⊵\mu$.

Diagonal: $d_{\mu \mu }=1$ for every $\mu$ with $D^{\mu }\mathrm{\ne }0$. The form radical gives the subquotient $D^{\mu }=C^{\mu }\mathrm{/}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$, and additivity of composition multiplicities [F4] gives $d_{\mu \mu }=[C^{\mu }:D^{\mu }]=[C^{\mu }\mathrm{/}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }:D^{\mu }]+[\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }:D^{\mu }]=1+[\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }:D^{\mu }]$, because $C^{\mu }\mathrm{/}\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }=D^{\mu }$ is simple by [F1]. The span of the $c_{uv}^{\mu }$ annihilates $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$: for $x\in \mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$ and all $u,v$ one has $x{c}_{uv}^{\mu }=⟨x,c_u^{\mu }{⟩}_{\mu }{c}_v^{\mu }=0$ by [F2]. Hence every element of this span acts as zero on $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$ and on all its subquotients, so $D^{\mu }$ is not a composition factor of $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }$; otherwise step 1.2 would produce an element of $\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{c_{uv}^{\mu }\}\subseteq A^{⊵\mu }$ acting nontrivially on it. Therefore $[\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\mu }:D^{\mu }]=0$ and $d_{\mu \mu }=1$.

Assembly: order the labels by a linear extension from smaller to larger, and restrict both rows and columns to labels with nonzero form quotient. Step 2.1 puts every nonzero entry on or below the diagonal of this square submatrix, and step 2.2 gives diagonal entries $1$. For the full rectangular matrix retain the label comparison $\lambda ⊵\mu$; deleting zero-quotient columns need not leave the labelled diagonal on the numerical matrix diagonal. These statements determine only the diagonal and the vanishing region.