Generally, vector space is an algebraic structure specified by the set of vectors $V$ and the set of scalars $\mathbb{F}$. So we concern about the homomorphism between two vector spaces in the normal sense - operation preservation. Concretely, if $f: V\to W$ is a homomorphism, then it preserves vector addition and scalar multiplication.$$f(x+y) = f(x)+f(y),\,\,\forall x,y\in V$$$$f(ax) = af(x),\,\,\forall x\in V, c\in\mathbb{F}$$The above properties are similar to that of linear function in analysis. So we use the term "linear transformation" interchangeably with "homomorphism" in the case of vector spaces.For example, let $P_3[\mathbb{R}]$ and $P_2[\mathbb{R}]$ be $\mathbb{R}$ vector spaces of polynomials of degree not exceeding 3 and 2, repectively. The derivative$$\dfrac{\mathrm{d}}{\mathrm{d}x} : P_3[\mathbb{R}] \to P_2[\mathbb{R}]$$ is a linear transformation.
Let $\beta=\{\beta_1,...,\beta_m\}$ and $\gamma=\{\gamma_1,...,\gamma_n\}$ are bases for $V$ and $W$, respectively. Then each $f(\beta_1),...,f(\beta_m)\in W$ can be express uniquely as linear combinations of $\gamma_1,...,\gamma_n$, concretely $$f(\beta_i) = \sum\limits_{j=1}^n a_{ij}\gamma_i,\,\,j=1,...,n.$$ In terms of matrix product,$$\left(\begin{array}{c} f(\beta_1) \\ \vdots \\ f(\beta_n) \\\end{array}\right) = \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \\ \end{array}\right) \left(\begin{array}{c} \gamma_1 \\ \vdots \\ \gamma_n \\\end{array}\right).$$ The matrix $(a_{ij})$ is called the matrix of $f$ relative to bases $\beta$ and $\gamma$, denoted by $[f]_\beta^\gamma$. If $\beta=\gamma$, we simply denote it $[f]_\beta$. For some vector $v = b_m\beta_m +...+b_m\beta_m\in V$, we have $$f(v) = \left(\begin{array}{ccc} b_1 & ... & b_m \\ \end{array}\right) \left(\begin{array}{c} f(\beta_1) \\ \vdots \\ f(\beta_m) \\\end{array}\right) = \left(\begin{array}{ccc} b_1 & ... & b_n \\ \end{array}\right) \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \\ \end{array}\right) \left(\begin{array}{c} \gamma_1 \\ \vdots \\ \gamma_n \\\end{array}\right).$$ Hence $[f(v)]_\gamma = [v]_\beta[f]_\beta^\gamma$. Proof. For each transformation, the matrix has been constructed uniquely as shown above. Now for each matrix $A = (a_{ij})\in \mathcal{M}_{m\times n}^\mathbb{F}$, define $f$ such that $$f(\beta_i) = \sum\limits_{j=1}^n a_{ij}\gamma_j,\,\, 1\le i \le m, 1\le j \le n.$$ We need to show that $f$ is unique. Suppose that there is $g$ such that $$g(\beta_i) = \sum\limits_{j=1}^n a_{ij}\gamma_j,\,\, 1\le i \le m, 1\le j \le n,$$ we can easily see that $f(v)=g(v),\,\,\forall v\in V$. Therefore, $\Phi$ is a bijection. $$$$ To show that $\Phi$ is an isomorphism, we have to point out that $$\Phi(cf+g) = c[f]_\beta^\gamma + [g]_\beta^\gamma.$$ Indeed, since $(cf+g)(v) = cf(v) + g(v)$, we have $$[(cf+g)(v)]_\gamma = c[f(v)]_\gamma + [g(v)]_\gamma = c[v]_\beta[f]_\beta^\gamma + [v]_\beta[g]_\beta^\gamma = [v]_\beta[cf]_\beta^\gamma + [v]_\beta[g]_\beta^\gamma = [v]_\beta([cf]_\beta^\gamma + [g]_\beta^\gamma) = [v]_\beta([cf + g])_\beta^\gamma.$$ Hence $[cf + g])_\beta^\gamma$ is the matrix of $cf+g$ relative to bases $\beta$ and $\gamma$.