[proofplan] We prove the result by expanding trace and matrix multiplication entrywise. The only algebraic input is commutativity of $R$, which lets us interchange the two scalar factors appearing in each product. First we prove the cyclic identity $\operatorname{tr}(XY)=\operatorname{tr}(YX)$ for arbitrary matrices $X,Y \in R^{n \times n}$, then apply it to $X=P^{-1}A$ and $Y=P$. Finally, the inverse relation $PP^{-1}=I_n$ reduces $APP^{-1}$ to $A$. [/proofplan] [step:Prove cyclicity of trace for two square matrices over $R$] Let $X=(x_{ij})_{1 \leq i,j \leq n} \in R^{n \times n}$ and $Y=(y_{ij})_{1 \leq i,j \leq n} \in R^{n \times n}$. By the definitions of matrix multiplication and trace, \begin{align*} \operatorname{tr}(XY)=\sum_{i=1}^{n}(XY)_{ii}=\sum_{i=1}^{n}\sum_{j=1}^{n}x_{ij}y_{ji}. \end{align*} Since $R$ is commutative, $x_{ij}y_{ji}=y_{ji}x_{ij}$ for every $i,j \in \{1,\dots,n\}$. Reindexing the finite sum by exchanging the names of $i$ and $j$ gives \begin{align*} \sum_{i=1}^{n}\sum_{j=1}^{n}x_{ij}y_{ji}=\sum_{j=1}^{n}\sum_{i=1}^{n}y_{ji}x_{ij}=\sum_{j=1}^{n}(YX)_{jj}=\operatorname{tr}(YX). \end{align*} Thus $\operatorname{tr}(XY)=\operatorname{tr}(YX)$. [guided] Let $X=(x_{ij})_{1 \leq i,j \leq n}$ and $Y=(y_{ij})_{1 \leq i,j \leq n}$ be arbitrary matrices in $R^{n \times n}$. We want to show that the trace does not see the order of the two factors in a product. Expanding the diagonal entries of $XY$ using the definition of matrix multiplication gives \begin{align*} (XY)_{ii}=\sum_{j=1}^{n}x_{ij}y_{ji} \end{align*} for each $i \in \{1,\dots,n\}$. Therefore the trace of $XY$ is \begin{align*} \operatorname{tr}(XY)=\sum_{i=1}^{n}(XY)_{ii}=\sum_{i=1}^{n}\sum_{j=1}^{n}x_{ij}y_{ji}. \end{align*} The key point is exactly where commutativity of $R$ is used. For each pair of indices $i,j$, the entries $x_{ij}$ and $y_{ji}$ are elements of the commutative ring $R$, so \begin{align*} x_{ij}y_{ji}=y_{ji}x_{ij}. \end{align*} Substituting this into the finite double sum gives \begin{align*} \operatorname{tr}(XY)=\sum_{i=1}^{n}\sum_{j=1}^{n}y_{ji}x_{ij}. \end{align*} Because the sum is finite, we may exchange the order of summation and rename the outer index as the diagonal index of $YX$: \begin{align*} \sum_{i=1}^{n}\sum_{j=1}^{n}y_{ji}x_{ij}=\sum_{j=1}^{n}\sum_{i=1}^{n}y_{ji}x_{ij}. \end{align*} For each fixed $j$, the inner sum is precisely the $j$th diagonal entry of $YX$: \begin{align*} (YX)_{jj}=\sum_{i=1}^{n}y_{ji}x_{ij}. \end{align*} Hence \begin{align*} \operatorname{tr}(XY)=\sum_{j=1}^{n}(YX)_{jj}=\operatorname{tr}(YX). \end{align*} This proves the cyclic identity $\operatorname{tr}(XY)=\operatorname{tr}(YX)$ for two square matrices over the commutative ring $R$. [/guided] [/step] [step:Apply cyclicity to move $P$ past $P^{-1}A$] Define $Q \in R^{n \times n}$ by $Q=P^{-1}$. Since $A,Q,P \in R^{n \times n}$, the product $QA \in R^{n \times n}$ is defined. Applying the cyclicity identity from the previous step with $X=QA$ and $Y=P$ gives \begin{align*} \operatorname{tr}(P^{-1}AP)=\operatorname{tr}((QA)P)=\operatorname{tr}(P(QA)). \end{align*} By associativity of matrix multiplication, \begin{align*} P(QA)=(PQ)A. \end{align*} [/step] [step:Use invertibility of $P$ to reduce the product to $A$] Since $Q=P^{-1}$ is the inverse of $P$, it satisfies $PQ=I_n$ and $QP=I_n$, where $I_n \in R^{n \times n}$ is the identity matrix. Therefore \begin{align*} P(QA)=(PQ)A=I_nA=A. \end{align*} Combining this with the equality from the previous step yields \begin{align*} \operatorname{tr}(P^{-1}AP)=\operatorname{tr}(A). \end{align*} This is the desired similarity invariance of the trace. [/step]