What can be said about the dimension of a direct sum? If \mathcal{U} is n -dimensional, \mathcal{V} is m -dimensional, and \mathcal{W} = \mathcal{U} \oplus \mathcal{V} , what is the dimension of \mathcal{W} ? This question is easy to answer.
Theorem 1. The dimension of a direct sum is the sum of the dimensions of its summands.
Proof. We assert that if \{x_1, \ldots, x_n\} is a basis in \mathcal{U} , and if \{y_1, \ldots, y_m\} is a basis in \mathcal{V} , then the set \{x_1, \ldots, x_n, y_1, \ldots, y_m\} (or, more precisely, the set \{\langle x_1, 0 \rangle, \ldots, \langle x_n, 0 \rangle, \langle 0, y_1 \rangle, \ldots, \langle 0, y_m \rangle\} ) is a basis in \mathcal{W} . The easiest proof of this assertion is to use the implication (1) \implies (3) from the theorem of the preceding section. Since every z in \mathcal{W} may be written in the form z = x + y , where x is a linear combination of x_1, \ldots, x_n and y is a linear combination of y_1, \ldots, y_m , it follows that our set does indeed span \mathcal{W} . To show that the set is also linearly independent, suppose that \alpha_1 x_1 + \cdots + \alpha_n x_n + \beta_1 y_1 + \cdots + \beta_m y_m = 0. The uniqueness of the representation of 0 in the form x + y implies that \alpha_1 x_1 + \cdots + \alpha_n x_n = \beta_1 y_1 + \cdots + \beta_m y_m = 0, and hence the linear independence of the x ’s and of the y ’s implies that \alpha_1 = \cdots = \alpha_n = \beta_1 = \cdots = \beta_m = 0. ◻
Theorem 2. If \mathcal{W} is any (n + m) -dimensional vector space, and if \mathcal{U} is any n -dimensional subspace of \mathcal{W} , then there exists an m -dimensional subspace \mathcal{V} in \mathcal{W} such that \mathcal{W} = \mathcal{U} \oplus \mathcal{V} .
Proof. Let \{x_1, \ldots, x_n\} be any basis in \mathcal{U} ; by the theorem of Section: Bases we may find a set \{y_1, \ldots, y_m\} of vectors in \mathcal{W} with the property that \{x_1, \ldots, x_n, y_1, \ldots, y_m\} is a basis in \mathcal{W} . Let \mathcal{V} be the subspace spanned by y_1, \ldots, y_m ; we omit the verification that \mathcal{W} = \mathcal{U} \oplus \mathcal{V} . ◻
Theorem 2 says that every subspace of a finite-dimensional vector space has a complement.
