[proofplan] We reduce the projective identity to the ordinary dimension formula for finite-dimensional vector subspaces. The only point requiring verification is the vector-space formula, which we prove directly by choosing a basis of $A \cap B$ and extending it to bases of $A$ and $B$. Once this formula is established, the convention $\dim \mathbb{P}(W)=\dim_K W-1$ for every nonzero finite-dimensional [vector space](/page/Vector%20Space) $W$ converts the desired equality into an immediate calculation. [/proofplan]

[step:Verify the involved projective spaces have defined dimensions] Since $A$ and $B$ are nonzero finite-dimensional $K$-vector spaces, $\mathbb{P}(A)$ and $\mathbb{P}(B)$ are defined and have dimensions \begin{align*} \dim \mathbb{P}(A) = \dim_K A - 1 \end{align*} and \begin{align*} \dim \mathbb{P}(B) = \dim_K B - 1. \end{align*} The hypothesis $A \cap B \neq \{0\}$ implies that $A \cap B$ is a nonzero finite-dimensional $K$-vector space, so \begin{align*} \dim \mathbb{P}(A \cap B) = \dim_K(A \cap B) - 1. \end{align*} Also $A+B$ is a finite-dimensional subspace of $V$ containing the nonzero subspace $A$, hence $A+B \neq \{0\}$, and therefore \begin{align*} \dim \mathbb{P}(A+B) = \dim_K(A+B) - 1. \end{align*} [/step]

[step:Prove the vector-space dimension formula by extending a basis]Let $r := \dim_K(A \cap B)$. Choose a basis \begin{align*} \mathcal{C} := \{c_1,\ldots,c_r\} \end{align*} of $A \cap B$. Since $A \cap B$ is a subspace of the finite-dimensional vector space $A$, extend $\mathcal{C}$ to a basis \begin{align*} \mathcal{A} := \{c_1,\ldots,c_r,a_1,\ldots,a_p\} \end{align*} of $A$. Since $A \cap B$ is also a subspace of the finite-dimensional vector space $B$, extend $\mathcal{C}$ to a basis \begin{align*} \mathcal{B} := \{c_1,\ldots,c_r,b_1,\ldots,b_q\} \end{align*} of $B$. We claim that \begin{align*} \mathcal{S} := \{c_1,\ldots,c_r,a_1,\ldots,a_p,b_1,\ldots,b_q\} \end{align*} is a basis of $A+B$. First, $\mathcal{S}$ spans $A+B$: if $x \in A+B$, then $x=u+v$ for some $u \in A$ and $v \in B$, and the basis expansions of $u$ in $\mathcal{A}$ and of $v$ in $\mathcal{B}$ express $x$ as a $K$-linear combination of the vectors in $\mathcal{S}$. It remains to prove [linear independence](/page/Linear%20Independence). Suppose \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{j=1}^{p}\beta_j a_j + \sum_{k=1}^{q}\gamma_k b_k = 0 \end{align*} with scalars $\alpha_i,\beta_j,\gamma_k \in K$. Then \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{j=1}^{p}\beta_j a_j = -\sum_{k=1}^{q}\gamma_k b_k. \end{align*} The left-hand side lies in $A$, and the right-hand side lies in $B$, so this common vector lies in $A \cap B$. Since $\mathcal{C}$ is a basis of $A \cap B$, there are scalars $\delta_1,\ldots,\delta_r \in K$ such that \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{j=1}^{p}\beta_j a_j = \sum_{i=1}^{r}\delta_i c_i. \end{align*} Rearranging gives \begin{align*} \sum_{i=1}^{r}(\alpha_i-\delta_i)c_i + \sum_{j=1}^{p}\beta_j a_j = 0. \end{align*} Since $\mathcal{A}$ is linearly independent, $\beta_j=0$ for every $1 \leq j \leq p$. Substituting this into the original relation gives \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{k=1}^{q}\gamma_k b_k = 0. \end{align*} Since $\mathcal{B}$ is linearly independent, $\alpha_i=0$ for every $1 \leq i \leq r$ and $\gamma_k=0$ for every $1 \leq k \leq q$. Thus $\mathcal{S}$ is linearly independent. Therefore $\mathcal{S}$ is a basis of $A+B$, and hence \begin{align*} \dim_K(A+B) = r+p+q. \end{align*} Also \begin{align*} \dim_K A = r+p \end{align*} and \begin{align*} \dim_K B = r+q. \end{align*} Substituting these identities yields \begin{align*} \dim_K(A+B) = \dim_K A + \dim_K B - \dim_K(A \cap B). \end{align*}[/step]

[guided]The goal of this step is to prove the ordinary dimension formula without treating it as a black box. Let \begin{align*} r := \dim_K(A \cap B). \end{align*} Because $A \cap B$ is nonzero and finite-dimensional, it has a basis, which we denote by \begin{align*} \mathcal{C} := \{c_1,\ldots,c_r\}. \end{align*} The reason to begin with a basis of the intersection is that the vectors common to $A$ and $B$ are exactly the vectors that would otherwise be counted twice. Since $A \cap B \subset A$ and $A$ is finite-dimensional, the basis-[extension theorem](/theorems/59) gives vectors $a_1,\ldots,a_p \in A$ such that \begin{align*} \mathcal{A} := \{c_1,\ldots,c_r,a_1,\ldots,a_p\} \end{align*} is a basis of $A$. Similarly, since $A \cap B \subset B$ and $B$ is finite-dimensional, there are vectors $b_1,\ldots,b_q \in B$ such that \begin{align*} \mathcal{B} := \{c_1,\ldots,c_r,b_1,\ldots,b_q\} \end{align*} is a basis of $B$. We now prove that the combined list \begin{align*} \mathcal{S} := \{c_1,\ldots,c_r,a_1,\ldots,a_p,b_1,\ldots,b_q\} \end{align*} is a basis of $A+B$. First, it spans $A+B$. Indeed, take any $x \in A+B$. By definition of the sum of subspaces, there exist $u \in A$ and $v \in B$ such that $x=u+v$. Since $\mathcal{A}$ is a basis of $A$, the vector $u$ is a $K$-linear combination of $c_1,\ldots,c_r,a_1,\ldots,a_p$. Since $\mathcal{B}$ is a basis of $B$, the vector $v$ is a $K$-linear combination of $c_1,\ldots,c_r,b_1,\ldots,b_q$. Adding these two expansions expresses $x$ as a $K$-linear combination of the vectors in $\mathcal{S}$. Next, we prove linear independence. Suppose scalars $\alpha_i,\beta_j,\gamma_k \in K$ satisfy \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{j=1}^{p}\beta_j a_j + \sum_{k=1}^{q}\gamma_k b_k = 0. \end{align*} Move the $B$-extension part to the other side: \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{j=1}^{p}\beta_j a_j = -\sum_{k=1}^{q}\gamma_k b_k. \end{align*} The left-hand side belongs to $A$ because all $c_i$ and $a_j$ lie in $A$. The right-hand side belongs to $B$ because all $b_k$ lie in $B$. Therefore the common vector lies in $A \cap B$. Since $\mathcal{C}$ is a basis of $A \cap B$, there exist scalars $\delta_1,\ldots,\delta_r \in K$ such that \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{j=1}^{p}\beta_j a_j = \sum_{i=1}^{r}\delta_i c_i. \end{align*} After rearranging, we obtain \begin{align*} \sum_{i=1}^{r}(\alpha_i-\delta_i)c_i + \sum_{j=1}^{p}\beta_j a_j = 0. \end{align*} This is a linear relation among the basis vectors in $\mathcal{A}$. Since $\mathcal{A}$ is linearly independent, every coefficient in this relation is zero, so in particular $\beta_j=0$ for every $1 \leq j \leq p$. Substituting $\beta_j=0$ into the original relation gives \begin{align*} \sum_{i=1}^{r}\alpha_i c_i + \sum_{k=1}^{q}\gamma_k b_k = 0. \end{align*} This is a linear relation among the basis vectors in $\mathcal{B}$. Since $\mathcal{B}$ is linearly independent, $\alpha_i=0$ for every $1 \leq i \leq r$ and $\gamma_k=0$ for every $1 \leq k \leq q$. Hence all coefficients in the original relation are zero, so $\mathcal{S}$ is linearly independent. Thus $\mathcal{S}$ is a basis of $A+B$. Counting the vectors in the three bases gives \begin{align*} \dim_K(A+B) = r+p+q. \end{align*} The bases $\mathcal{A}$ and $\mathcal{B}$ give \begin{align*} \dim_K A = r+p \end{align*} and \begin{align*} \dim_K B = r+q. \end{align*} Since $r=\dim_K(A \cap B)$, these three equalities imply \begin{align*} \dim_K(A+B) = \dim_K A + \dim_K B - \dim_K(A \cap B). \end{align*} This is the exact vector-space identity needed to transfer the calculation to projective dimensions.[/guided]

[step:Translate the vector-space formula into projective dimensions] Using $\dim \mathbb{P}(W)=\dim_K W-1$ for each nonzero finite-dimensional vector space $W$ in question, we compute \begin{align*} \dim \mathbb{P}(A) + \dim \mathbb{P}(B) = (\dim_K A - 1) + (\dim_K B - 1). \end{align*} Thus \begin{align*} \dim \mathbb{P}(A) + \dim \mathbb{P}(B) = \dim_K A + \dim_K B - 2. \end{align*} On the other hand, \begin{align*} \dim \mathbb{P}(A+B) + \dim \mathbb{P}(A \cap B) = (\dim_K(A+B)-1) + (\dim_K(A \cap B)-1). \end{align*} By the vector-space dimension formula proved above, \begin{align*} \dim_K(A+B)+\dim_K(A \cap B) = \dim_K A + \dim_K B. \end{align*} Therefore \begin{align*} \dim \mathbb{P}(A+B) + \dim \mathbb{P}(A \cap B) = \dim_K A + \dim_K B - 2. \end{align*} Both sides are equal to $\dim_K A+\dim_K B-2$, so \begin{align*} \dim \mathbb{P}(A) + \dim \mathbb{P}(B) = \dim \mathbb{P}(A+B) + \dim \mathbb{P}(A \cap B). \end{align*} This proves the claimed projective dimension formula. [/step]