k*G^r \to k*G^s\bigr) \bigr),
\label{dim_Skg_is_dim}$$ where $r_A$ denotes the module homomorphism given by right multiplication with $A$. First note that we may assume that $r=s$. Indeed if $r<s$, replace $A$ with the $s \times s$ matrix which is $A$ for the first $r$ rows, and has 0’s on the bottom $s-r$ rows. On the other hand if $r>s$, replace $A$ with the $r \times r$ matrix $B$ with entries $(b_{ij})$ which is $A$ for the first $s$ columns, and has $b_{ij} =
\delta_{ij}$ if $i>s$, where $\delta_{ij}$ is the Kronecker delta.

We will often use the obvious long exact sequence associated to homomorphisms $f \colon M_0 \to M_1$ and $g \colon
M_1 \to M_2$ $$0 \to \ker(f) \to \ker(g\circ f) \to \ker(g) 
\to {\operatorname{coker}}(f) \to {\operatorname{coker}}(g\circ f) \to {\operatorname{coker}}(g) \to 0.
\label{ker-coker-sequence}$$ We now assume that $A$ is an $r \times r$ matrix. Note that equation is true if $A$ is invertible over $S^{-1}k*G$; this is because then $\ker r_A = 0$ (whether $A$ is considered as a matrix over $k*G$ or $S^{-1}k*G$).

Next observe that if $U \in \operatorname{M}_r(k*G)$ which is invertible over $\operatorname{M}_r(S^{-1}k*G)$, then equation holds for $A$ if and only if it holds for $AU$, and also 