Let \(V\) be a finite-dimensional vector space and let \(W_1, \ldots, W_k\) be subspaces of \(V\) such that \[ V = W_1 + \cdots + W_k \quad \text{and} \quad \dim V = \dim W_1 + \cdots + \dim W_k. \] Prove that \(V = W_1 \oplus \cdots \oplus W_k\).
Solution
1. The union of bases of the \(W_i\) is a basis of \(V\).
For each \(i\), choose a basis \(\mathcal B_i\) of \(W_i\). Since \[ V=W_1+\cdots+W_k, \] the concatenated list \[ \mathcal B_1,\ldots,\mathcal B_k \] spans \(V\). Its total number of vectors is \[ \sum_{i=1}^k|\mathcal B_i| =\sum_{i=1}^k\dim W_i =\dim V. \] A spanning list in a finite-dimensional vector space whose length equals the dimension is a basis [@HK71]. Hence the concatenated list is linearly independent.
2. Linear independence gives uniqueness of the decomposition.
1.Linear independence forces every coefficient to vanish. Hence\[ w_1=\cdots=w_k=0. \]Therefore the representation of a vector as a sum of vectors from the \(W_i\)is unique. Together with the given equality \(V=W_1+\cdots+W_k\), this isexactly\[ \boxed{V=W_1\oplus\cdots\oplus W_k}. \]