[proofplan] We start with a basis for $U$, extend it to a basis for $V$ using [Properties of Finite Dimensional Bases](/theorems/374) part (v), and show that the images of the extension vectors under the quotient map $\pi: V \to V/U$ form a basis for $V/U$. The spanning property follows from applying $\pi$ to the basis expansion (the $U$-components vanish), and independence follows from the independence of the full basis for $V$. [/proofplan] [step:Extend a basis for $U$ to a basis for $V$] Let $\dim U = k$ and $\dim V = n$. Choose a basis $\{u_1, \ldots, u_k\}$ for $U$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v), extend to a basis $\{u_1, \ldots, u_k, v_{k+1}, \ldots, v_n\}$ for $V$. Let $\pi: V \to V/U$ denote the canonical quotient map $\pi(v) = v + U$. [/step] [step:Show the images $\{\pi(v_{k+1}), \ldots, \pi(v_n)\}$ span $V/U$] Let $v + U \in V/U$. Since $\{u_1, \ldots, u_k, v_{k+1}, \ldots, v_n\}$ is a basis for $V$, there exist scalars $\alpha_1, \ldots, \alpha_k, \lambda_{k+1}, \ldots, \lambda_n \in \mathbb{F}$ with \begin{align*} v = \sum_{i=1}^{k}\alpha_i u_i + \sum_{j=k+1}^{n}\lambda_j v_j. \end{align*} Applying $\pi$ and using $\pi(u_i) = \mathbf{0}_{V/U}$ (since $u_i \in U$): \begin{align*} v + U = \sum_{j=k+1}^{n}\lambda_j(v_j + U). \end{align*} Hence $\{\pi(v_{k+1}), \ldots, \pi(v_n)\}$ spans $V/U$. [/step] [step:Show the images $\{\pi(v_{k+1}), \ldots, \pi(v_n)\}$ are linearly independent] Suppose $\sum_{j=k+1}^{n}\lambda_j(v_j + U) = U$ (the zero element of $V/U$). Then $\sum_{j=k+1}^{n}\lambda_j v_j \in U$, so there exist $\alpha_1, \ldots, \alpha_k \in \mathbb{F}$ with \begin{align*} \sum_{j=k+1}^{n}\lambda_j v_j = \sum_{i=1}^{k}\alpha_i u_i. \end{align*} Rearranging: $\sum_{i=1}^{k}(-\alpha_i)u_i + \sum_{j=k+1}^{n}\lambda_j v_j = \mathbf{0}$. Since $\{u_1, \ldots, u_k, v_{k+1}, \ldots, v_n\}$ is a basis for $V$ (hence linearly independent), all coefficients vanish. In particular, $\lambda_j = 0$ for all $j \in \{k+1, \ldots, n\}$. [/step] [step:Conclude the dimension formula] The set $\{\pi(v_{k+1}), \ldots, \pi(v_n)\}$ is a basis for $V/U$ with $n - k$ elements: \begin{align*} \dim(V/U) = n - k = \dim V - \dim U. \end{align*} Rearranging gives $\dim V = \dim U + \dim(V/U)$. [/step]