> [!NOTE] Corollary (Commutative square) > Let $V$ be a FDVS with basis $B=\{ v_{1},\dots,v_{n} \}$ and $W$ be a FDVS with basis $B' = \{ w_{1}, \dots , w_{m} \}$. If $\varphi: V\to W$ is a linear map, then $\varphi = \varphi_{A}$ for some matrix $A \in \mathbb{F}^{mn}$ and there is a [[Commutative Square|commutative square]] of linear maps $\begin{CD} V @> \varphi_{A} >> W\\ @VV\chi_{B} V @VV \chi_{B'} V\\ \mathbb{R}^{n} @>L_{A}>> \mathbb{R}^{m} \end{CD}$ >*Proof*. The maps are well defined: $\varphi_{A}$ is the [[Left Multiplication Linear Map of Real Matrix#^2b50ef|left multiplication linear map corresponding to A wrt to bases of V and W]]. $L_{A}$ is the [[Left Multiplication Linear Map of Real Matrix#^2488d3|left multiplication linear map]] of $A$. $\chi_{B}$ and $\chi_{B'}$ are the [[Linear Isomorphism#^105ca6|linear isomorphisms]] from $V$ and $W$ to $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ respectively. >The claim is simply that $\chi_{B'} \circ \varphi_{A} = L_{A} \circ \chi_{B}.$Let $v\in V$. Then $\exists! \lambda_{i} \in \mathbb{F}$ such that $v=\lambda_{1}v_{1}+\dots\lambda_{n}v_{n}$. On the one hand, $\chi_{B}(v)=(\lambda_{1},\dots,\lambda_{n})^{T} \implies L_{A}(\chi_{B}(v)) = A \, (\lambda_{1},\dots,\lambda_{n})^{T}$On the other hand, $\varphi_{A}(v) = \mu_{1}w_{1}+\dots+\mu_{m}w_{m}$where the coefficients $\mu_{j}$ of the image are defined by $(\mu_{1},\dots,\mu_{m})^T =A \, (\lambda_{1},\dots,\lambda_{n})^{T}$and so $\chi_{B'}(\varphi_{A}) = (\mu_{1},\dots,\mu_{m})^{T } = A \, (\lambda_{1},\dots,\lambda_{n})^{T}$These two are equal as claimed.