[proofplan] We compare the two matrices by writing the second as the first conjugated on the left and right by change-of-basis matrices. The change-of-basis matrices are invertible, so left multiplication changes only the coordinate description of the range, and right multiplication only reparametrizes the domain. Therefore neither operation changes the dimension of the image, which is the rank. [/proofplan] [step:Express the two representing matrices using change of basis matrices] Let $n=\dim V$ and $m=\dim W$. Define \begin{align*}A := [T]_{\mathcal D \leftarrow \mathcal B} \in k^{m\times n}\end{align*} and \begin{align*} A' := [T]_{\mathcal E \leftarrow \mathcal C} \in k^{m\times n}. \end{align*} Let \begin{align*}P := P_{\mathcal E \leftarrow \mathcal D} \in k^{m\times m}\end{align*} be the change-of-basis matrix from $\mathcal D$-coordinates to $\mathcal E$-coordinates, and let \begin{align*}Q := P_{\mathcal B \leftarrow \mathcal C} \in k^{n\times n}\end{align*} be the change-of-basis matrix from $\mathcal C$-coordinates to $\mathcal B$-coordinates. For every $v\in V$, the defining property of $A$ gives \begin{align*}[T(v)]_{\mathcal D}=A[v]_{\mathcal B}.\end{align*} Since $P$ converts $\mathcal D$-coordinates to $\mathcal E$-coordinates and $Q$ converts $\mathcal C$-coordinates to $\mathcal B$-coordinates, we have \begin{align*}[T(v)]_{\mathcal E}=P[T(v)]_{\mathcal D}=PA[v]_{\mathcal B}=PAQ[v]_{\mathcal C}.\end{align*} By the defining property of the representing matrix $[T]_{\mathcal E \leftarrow \mathcal C}$, this implies \begin{align*}A'=PAQ.\end{align*} By [citetheorem:8324], the change-of-basis matrices $P$ and $Q$ are invertible. [/step] [step:Show that multiplying by invertible matrices preserves rank] We prove the elementary rank fact needed in this setting. Let $R\in k^{m\times m}$ be invertible and let $S\in k^{n\times n}$ be invertible. Let $M\in k^{m\times n}$ be any matrix. Regard $M$ as the [linear map](/page/Linear%20Map) $M:k^n\to k^m$ given by column multiplication. Right multiplication by $S$ does not change the image, because $S:k^n\to k^n$ is surjective. Hence \begin{align*} \operatorname{im}(MS)=M(S(k^n))=M(k^n)=\operatorname{im}(M). \end{align*} Therefore \begin{align*} \operatorname{rank}(MS)=\operatorname{rank}(M). \end{align*} Left multiplication by $R$ sends $\operatorname{im}(M)$ isomorphically onto $\operatorname{im}(RM)$. Indeed, \begin{align*} \operatorname{im}(RM)=R(\operatorname{im}(M)). \end{align*} Since $R:k^m\to k^m$ is an isomorphism, its restriction to $\operatorname{im}(M)$ is an isomorphism from $\operatorname{im}(M)$ onto $R(\operatorname{im}(M))$. Thus \begin{align*} \operatorname{rank}(RM)=\operatorname{rank}(M). \end{align*} Applying these two conclusions successively gives \begin{align*} \operatorname{rank}(RMS)=\operatorname{rank}(M). \end{align*} [guided] We need only one general fact: invertible changes of coordinates do not alter the dimension of a matrix image. Let $R\in k^{m\times m}$ and $S\in k^{n\times n}$ be invertible matrices, and let $M\in k^{m\times n}$. We view $M$ as the linear map $M:k^n\to k^m$ defined by multiplying column vectors. First consider the right multiplication $MS$. The matrix $S$ acts before $M$. Since $S$ is invertible, the map $S:k^n\to k^n$ is surjective, so its image is all of $k^n$. Therefore \begin{align*} \operatorname{im}(MS)=M(S(k^n))=M(k^n)=\operatorname{im}(M). \end{align*} The rank of a matrix is the dimension of its image, so \begin{align*} \operatorname{rank}(MS)=\operatorname{rank}(M). \end{align*} Now consider left multiplication by $R$. The image of $RM$ is obtained by applying $R$ to every vector in the image of $M$: \begin{align*} \operatorname{im}(RM)=R(\operatorname{im}(M)). \end{align*} Because $R:k^m\to k^m$ is an isomorphism, its restriction to the subspace $\operatorname{im}(M)\subset k^m$ is an isomorphism from $\operatorname{im}(M)$ onto $R(\operatorname{im}(M))$. Isomorphic finite-dimensional vector spaces have the same dimension, so \begin{align*} \operatorname{rank}(RM)=\operatorname{rank}(M). \end{align*} Combining the right-multiplication and left-multiplication arguments yields \begin{align*} \operatorname{rank}(RMS)=\operatorname{rank}(M). \end{align*} [/guided] [/step] [step:Apply rank invariance to the change of basis formula] From the first step, we have \begin{align*} A'=PAQ, \end{align*} where $P$ and $Q$ are invertible. Applying the rank invariance result from the previous step with $R=P$, $M=A$, and $S=Q$, we obtain \begin{align*}\operatorname{rank}(A')=\operatorname{rank}(PAQ)=\operatorname{rank}(A).\end{align*} Substituting back the definitions of $A$ and $A'$ gives \begin{align*}\operatorname{rank}([T]_{\mathcal E \leftarrow \mathcal C})=\operatorname{rank}([T]_{\mathcal D \leftarrow \mathcal B}).\end{align*} This is the desired equality. [/step]