[proofplan] We construct the complement $W$ by extending a basis for $U$ to a basis for $V$ using the [Extension of Independent Sets to Bases](/theorems/3270), then taking $W$ to be the span of the appended vectors. We verify the direct sum conditions $U + W = V$ and $U \cap W = \{0\}$ directly from the basis structure. The dimension formula then follows from the [Dimension of a Direct Sum](/theorems/3273). [/proofplan] [step:Extend a basis for $U$ to a basis for $V$] Let $k := \dim U$ and $n := \dim V$. Since $U$ is a subspace of the finite-dimensional space $V$, we have $k \le n$. Let $\{u_1, \ldots, u_k\}$ be a basis for $U$. Since this is a linearly independent subset of $V$ with $k \le n$ elements, the [Extension of Independent Sets to Bases](/theorems/3270) provides vectors $v_{k+1}, \ldots, v_n \in V$ such that \begin{align*} \mathcal{B} := \{u_1, \ldots, u_k, v_{k+1}, \ldots, v_n\} \end{align*} is a basis for $V$. [/step] [step:Define $W = \operatorname{span}(v_{k+1}, \ldots, v_n)$ and verify $V = U + W$] Define the subspace \begin{align*} W := \operatorname{span}(v_{k+1}, \ldots, v_n). \end{align*} We show $V = U + W$. Let $v \in V$. Since $\mathcal{B}$ is a basis for $V$, there exist scalars $a_1, \ldots, a_k, b_{k+1}, \ldots, b_n$ with \begin{align*} v = a_1 u_1 + \cdots + a_k u_k + b_{k+1} v_{k+1} + \cdots + b_n v_n. \end{align*} The first sum $a_1 u_1 + \cdots + a_k u_k$ lies in $U$ (as a linear combination of basis vectors of $U$), and the second sum $b_{k+1} v_{k+1} + \cdots + b_n v_n$ lies in $W$ (by definition of $W$). Therefore $v \in U + W$, and since $v$ was arbitrary, $V = U + W$. [/step] [step:Verify $U \cap W = \{0\}$ to establish the direct sum] Let $z \in U \cap W$. Since $z \in U$, there exist scalars $a_1, \ldots, a_k$ with $z = a_1 u_1 + \cdots + a_k u_k$. Since $z \in W$, there exist scalars $b_{k+1}, \ldots, b_n$ with $z = b_{k+1} v_{k+1} + \cdots + b_n v_n$. Equating: \begin{align*} a_1 u_1 + \cdots + a_k u_k - b_{k+1} v_{k+1} - \cdots - b_n v_n = 0. \end{align*} Since $\mathcal{B} = \{u_1, \ldots, u_k, v_{k+1}, \ldots, v_n\}$ is a basis for $V$, it is linearly independent. Therefore $a_1 = \cdots = a_k = 0$ and $b_{k+1} = \cdots = b_n = 0$, giving $z = 0$. We have shown $U \cap W = \{0\}$. Combined with $V = U + W$, this gives $V = U \oplus W$. [guided] The argument exploits the fact that the vectors $u_1, \ldots, u_k, v_{k+1}, \ldots, v_n$ together form a basis $\mathcal{B}$ for $V$, and a basis is in particular linearly independent. Any vector $z \in U \cap W$ has two representations: one as a combination of the $u_i$ (because $z \in U$ and $\{u_1, \ldots, u_k\}$ is a basis for $U$) and one as a combination of the $v_j$ (because $z \in W = \operatorname{span}(v_{k+1}, \ldots, v_n)$). Subtracting these two expressions: \begin{align*} a_1 u_1 + \cdots + a_k u_k - b_{k+1} v_{k+1} - \cdots - b_n v_n = 0. \end{align*} This is a linear combination of the $n$ vectors in $\mathcal{B}$ that equals zero. The linear independence of $\mathcal{B}$ forces every coefficient to vanish, so $z = 0$. This is the only place where the linear independence of the full basis $\mathcal{B}$ is used -- the spanning property was consumed in the previous step. [/guided] [/step] [step:Derive the dimension formula for the complement] Since $V = U \oplus W$, the [Dimension of a Direct Sum](/theorems/3273) gives \begin{align*} \dim V = \dim U + \dim W. \end{align*} Rearranging: $\dim W = \dim V - \dim U = n - k$. This formula holds for any complement $W$ of $U$, not just the one constructed above: if $V = U \oplus W'$ for any subspace $W'$, then the [Dimension of a Direct Sum](/theorems/3273) applied to the decomposition $V = U \oplus W'$ yields $\dim W' = \dim V - \dim U$. [/step]