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 \[\alpha_1 = \cdots = \alpha_n = \beta_1 = \cdots = \beta_m = 0.\] ◻

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 \(\mathcal{W} = \mathcal{U} \oplus \mathcal{V}\) . ◻