[proofplan] We construct a map $\Phi: \mathrm{Mat}_{n,m}(\mathbb{F}) \to \mathcal{L}(U, V)$ that sends each matrix $A$ to the unique linear map determined by its columns (via [Extension from Basis](/theorems/381)), and an inverse $\Psi$ that reads off coordinates of the images of basis vectors (via [Unique Representation by a Basis](/theorems/372)). We verify these are mutual inverses to establish bijectivity, then show $\Phi$ is $\mathbb{F}$-linear, making it a vector space isomorphism by the [Isomorphism iff Linear Bijection](/theorems/378) theorem. [/proofplan] [step:Define $\Phi$ from matrices to linear maps via column-to-image assignment] Fix an ordered basis $(e_1, \dots, e_m)$ for $U$ and an ordered basis $(f_1, \dots, f_n)$ for $V$. Given a matrix $A = (a_{ji}) \in \mathrm{Mat}_{n,m}(\mathbb{F})$, define vectors $v_i = \sum_{j=1}^{n} a_{ji} f_j \in V$ for each $i \in \{1, \dots, m\}$. By [Extension from Basis](/theorems/381), there exists a unique [linear map](/page/Linear%20Map) $\alpha: U \to V$ with $\alpha(e_i) = v_i$ for each $i$. Set $\Phi(A) = \alpha$. [guided] The idea is to interpret a matrix as a recipe for building a linear map: the $i$th column of $A$ contains the coordinates of the image $\alpha(e_i)$ in the basis $(f_1, \dots, f_n)$. Explicitly, for $A = (a_{ji}) \in \mathrm{Mat}_{n,m}(\mathbb{F})$, the $i$th column $(a_{1i}, \dots, a_{ni})^\top$ encodes the vector $v_i = \sum_{j=1}^{n} a_{ji} f_j \in V$. The entry $a_{ji}$ is the $j$th coordinate of the image of the $i$th basis vector. By [Extension from Basis](/theorems/381), there exists a unique linear map $\alpha: U \to V$ satisfying $\alpha(e_i) = v_i$ for each $i \in \{1, \dots, m\}$. We define $\Phi(A) = \alpha$. The existence and uniqueness guaranteed by the extension theorem ensure that $\Phi$ is a well-defined function from $\mathrm{Mat}_{n,m}(\mathbb{F})$ to $\mathcal{L}(U, V)$. For a concrete example: if $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ with $m = n = 2$, then $\alpha(e_1) = 3f_1 + 0f_2 = 3f_1$ and $\alpha(e_2) = 1f_1 + 2f_2$. For any $u = \lambda_1 e_1 + \lambda_2 e_2$, linearity gives $\alpha(u) = \lambda_1 \alpha(e_1) + \lambda_2 \alpha(e_2) = (3\lambda_1 + \lambda_2)f_1 + 2\lambda_2 f_2$, which is the standard matrix-vector product $A(\lambda_1, \lambda_2)^\top$ expressed in the basis $(f_j)$. [/guided] [/step] [step:Define the inverse $\Psi$ by reading off coordinates of images] Given $\alpha \in \mathcal{L}(U, V)$, for each $i \in \{1, \dots, m\}$ the vector $\alpha(e_i) \in V$ has a unique coordinate representation $\alpha(e_i) = \sum_{j=1}^{n} a_{ji} f_j$ by [Unique Representation by a Basis](/theorems/372). Define $\Psi(\alpha) = (a_{ji}) \in \mathrm{Mat}_{n,m}(\mathbb{F})$. [/step] [step:Verify $\Phi$ and $\Psi$ are mutual inverses] For $A \in \mathrm{Mat}_{n,m}(\mathbb{F})$: $\Psi(\Phi(A))$ reads off the coordinates of $\Phi(A)(e_i) = \sum_{j=1}^{n} a_{ji} f_j$ in the basis $(f_j)$, recovering the entries $a_{ji}$, so $\Psi(\Phi(A)) = A$. For $\alpha \in \mathcal{L}(U, V)$: $\Phi(\Psi(\alpha))$ is the unique linear map sending $e_i$ to $\sum_{j=1}^{n} a_{ji} f_j = \alpha(e_i)$. By the uniqueness statement in [Extension from Basis](/theorems/381), $\Phi(\Psi(\alpha)) = \alpha$. Hence $\Psi \circ \Phi = \mathrm{id}$ and $\Phi \circ \Psi = \mathrm{id}$, so $\Phi$ is a bijection. [/step] [step:Show $\Phi$ is $\mathbb{F}$-linear] Let $A, B \in \mathrm{Mat}_{n,m}(\mathbb{F})$ and $\lambda, \mu \in \mathbb{F}$. For each $i \in \{1, \dots, m\}$: \begin{align*} \Phi(\lambda A + \mu B)(e_i) &= \sum_{j=1}^{n} (\lambda a_{ji} + \mu b_{ji}) f_j = \lambda \sum_{j=1}^{n} a_{ji} f_j + \mu \sum_{j=1}^{n} b_{ji} f_j = \lambda \Phi(A)(e_i) + \mu \Phi(B)(e_i) \\ &= (\lambda \Phi(A) + \mu \Phi(B))(e_i). \end{align*} Since $\Phi(\lambda A + \mu B)$ and $\lambda \Phi(A) + \mu \Phi(B)$ are both linear maps that agree on the basis $(e_1, \dots, e_m)$, they are equal by uniqueness of linear extension ([Extension from Basis](/theorems/381)). [guided] We must verify $\Phi(\lambda A + \mu B) = \lambda \Phi(A) + \mu \Phi(B)$. Two linear maps from $U$ to $V$ are equal if and only if they agree on a basis for $U$ (by the uniqueness part of [Extension from Basis](/theorems/381)). So it suffices to check agreement on each $e_i$. The $i$th column of $\lambda A + \mu B$ has entries $\lambda a_{ji} + \mu b_{ji}$, so \begin{align*} \Phi(\lambda A + \mu B)(e_i) &= \sum_{j=1}^{n} (\lambda a_{ji} + \mu b_{ji}) f_j = \lambda \sum_{j=1}^{n} a_{ji} f_j + \mu \sum_{j=1}^{n} b_{ji} f_j \\ &= \lambda \Phi(A)(e_i) + \mu \Phi(B)(e_i) = (\lambda \Phi(A) + \mu \Phi(B))(e_i). \end{align*} Since this holds for every basis vector, $\Phi(\lambda A + \mu B) = \lambda \Phi(A) + \mu \Phi(B)$. [/guided] [/step] [step:Conclude that $\Phi$ is an isomorphism and compute $\dim \mathcal{L}(U, V)$] Since $\Phi$ is a bijective $\mathbb{F}$-linear map, it is a vector space isomorphism by the [Isomorphism iff Linear Bijection](/theorems/378) theorem. In particular, \begin{align*} \dim \mathcal{L}(U, V) = \dim \mathrm{Mat}_{n,m}(\mathbb{F}) = nm = (\dim U)(\dim V). \end{align*} [/step]