[proofplan] We start with a basis for $U \cap W$, extend it separately to bases for $U$ and for $W$ using [Properties of Finite Dimensional Bases](/theorems/374) part (v), then show the combined set is a basis for $U + W$. Counting the elements gives the dimension formula. [/proofplan] [step:Extend a basis for $U \cap W$ to bases for $U$ and $W$] Let $r = \dim(U \cap W)$ and choose a basis $\{v_1, \ldots, v_r\}$ for $U \cap W$. Since $U \cap W \subset U$, this is a linearly independent subset of $U$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v), extend it to a basis \begin{align*} \{v_1, \ldots, v_r, u_1, \ldots, u_s\} \end{align*} for $U$, so $\dim U = r + s$. Similarly, extend $\{v_1, \ldots, v_r\}$ to a basis \begin{align*} \{v_1, \ldots, v_r, w_1, \ldots, w_t\} \end{align*} for $W$, so $\dim W = r + t$. [/step] [step:Show $B = \{v_1, \ldots, v_r, u_1, \ldots, u_s, w_1, \ldots, w_t\}$ spans $U + W$] Any element of $U + W$ has the form $x + y$ with $x \in U$ and $y \in W$. Since $\{v_1, \ldots, v_r, u_1, \ldots, u_s\}$ is a basis for $U$, the vector $x$ is a linear combination of these. Since $\{v_1, \ldots, v_r, w_1, \ldots, w_t\}$ is a basis for $W$, the vector $y$ is a linear combination of these. Hence $x + y \in \langle B \rangle$, so $B$ spans $U + W$. [/step] [step:Show $B$ is linearly independent] Suppose \begin{align*} \sum_{i=1}^{r}\alpha_i v_i + \sum_{j=1}^{s}\beta_j u_j + \sum_{k=1}^{t}\gamma_k w_k = \mathbf{0}. \end{align*} Rearranging: \begin{align*} \sum_{k=1}^{t}\gamma_k w_k = -\sum_{i=1}^{r}\alpha_i v_i - \sum_{j=1}^{s}\beta_j u_j. \end{align*} The left side belongs to $W$ and the right side belongs to $U$, so both sides lie in $U \cap W$. Since $\{v_1, \ldots, v_r\}$ is a basis for $U \cap W$, there exist $\delta_1, \ldots, \delta_r \in \mathbb{F}$ with \begin{align*} \sum_{k=1}^{t}\gamma_k w_k = \sum_{i=1}^{r}\delta_i v_i. \end{align*} Rewriting: $\sum_{k=1}^{t}\gamma_k w_k - \sum_{i=1}^{r}\delta_i v_i = \mathbf{0}$. Since $\{v_1, \ldots, v_r, w_1, \ldots, w_t\}$ is a basis for $W$ (hence linearly independent), all coefficients vanish: $\gamma_k = 0$ for all $k$ and $\delta_i = 0$ for all $i$. Substituting $\gamma_k = 0$ back into the original equation: $\sum_{i=1}^{r}\alpha_i v_i + \sum_{j=1}^{s}\beta_j u_j = \mathbf{0}$. Since $\{v_1, \ldots, v_r, u_1, \ldots, u_s\}$ is a basis for $U$ (hence linearly independent), $\alpha_i = 0$ for all $i$ and $\beta_j = 0$ for all $j$. [guided] The independence proof hinges on one key idea: the "cross term" between the $U$-vectors and $W$-vectors must pass through $U \cap W$, where we already have a basis. Starting from $\sum \alpha_i v_i + \sum \beta_j u_j + \sum \gamma_k w_k = \mathbf{0}$, we isolate the $W$-specific vectors on one side: \begin{align*} \sum_{k=1}^{t}\gamma_k w_k = -\sum_{i=1}^{r}\alpha_i v_i - \sum_{j=1}^{s}\beta_j u_j. \end{align*} The left side is in $W$ (as a combination of $w_k$ and $v_i$ are in $W$ too, but we have only $w_k$ here). The right side is in $U$ (as a combination of $v_i$ and $u_j$, all of which lie in $U$). So both sides lie in $U \cap W = \langle v_1, \ldots, v_r \rangle$. Expressing the left side in terms of the basis $\{v_1, \ldots, v_r\}$ for $U \cap W$ and using the independence of the basis $\{v_1, \ldots, v_r, w_1, \ldots, w_t\}$ for $W$ forces $\gamma_k = 0$ for all $k$. Once the $\gamma_k$ vanish, the original equation reduces to a relation in $U$, and the independence of the $U$-basis forces $\alpha_i = 0$ and $\beta_j = 0$. [/guided] [/step] [step:Count elements to obtain the dimension formula] Since $B$ is a basis for $U + W$ with $r + s + t$ elements: \begin{align*} \dim(U + W) &= r + s + t \\ &= (r + s) + (r + t) - r \\ &= \dim U + \dim W - \dim(U \cap W). \end{align*} [/step]