Let \(W_1\) and \(W_2\) be subspaces of a vector space \(V\) such that \(W_1 + W_2 = V\) and \(W_1 \cap W_2 = \{0\}\). Prove that for each vector \(\alpha\) in \(V\) there are unique vectors \(\alpha_1\) in \(W_1\) and \(\alpha_2\) in \(W_2\) such that \(\alpha = \alpha_1 + \alpha_2\).
Solution
Existence follows from \(W_1+W_2=V\): for every \(\alpha\in V\) there are \(\alpha_1\in W_1\) and \(\alpha_2\in W_2\) with \[ \alpha=\alpha_1+\alpha_2. \] For uniqueness, suppose also \[ \alpha=\beta_1+\beta_2, \qquad \beta_1\in W_1, \quad \beta_2\in W_2. \] Then \[ \alpha_1-\beta_1=\beta_2-\alpha_2. \] The left side lies in \(W_1\) and the right side lies in \(W_2\), so their common value lies in \(W_1\cap W_2=\{0\}\). Hence \[ \alpha_1=\beta_1, \qquad \alpha_2=\beta_2. \] Thus the decomposition is unique.