Let $V$ be a finite-dimensional vector space and let $W_1,…,W_k$ be subspaces of $V$ such that $V=W_1+…+W_k$ and $\dim(V)=\dim(W_1)+…+\dim (W_k)$. Prove that $V=W_1\oplus …\oplus W_k$.
Approach (1): Suppose $V=W_1+…+W_k$ and $\dim(V)=\sum_{i=1}^k \dim (W_i)$. Then $\dim (V)=\dim (W_1+…+W_k)= \sum_{i=1}^k \dim (W_i)$. By theorem 6 section 2.3, $\dim (W_1+…+W_k)=\dim (W_1)+\dim (\sum_{i\gt 1}W_i)-\dim (W_1\cap \sum_{i\gt 1}W_i)$. By mathematical induction, $$\dim (W_1+…+W_k)=\sum_{i=1}^k \dim (W_i)-\sum_{j=1}^{k-1} \dim(W_j\cap \sum_{i\gt j}W_i).$$ So $\sum_{j=1}^{k-1} \dim(W_j\cap \sum_{i\gt j}W_i)=0$. Since $\dim(W_j\cap \sum_{i\gt j}W_i) \geq 0$, we have $\dim (W_j\cap \sum_{i\gt j}W_i)=0$. So $W_j\cap \sum_{i\gt j}W_i=\{0_V\}$, for all $1\leq j\leq k-1$. By lemma 1 section 6.6, $W_1,…,W_k$ are independent. Thus $V=W_1\oplus … \oplus W_k$. Is my proof correct?
Approach (2): Let $B_i$ be basis of $W_i$, $\forall i\in J_k$. Let $B=\bigcup_{i=1}^k B_i$. We show $B$ is basis of $V$. Since $V=W_1+…+W_k$, we have $V=\text{span} (B)$. Since $\dim (V)=\dim(W_1)+…+\dim (W_k)=n$, we have $\dim(W_1)+…+\dim (W_k)=|B_1|+…+|B_k|$. How to show $B_i\cap B_j=\emptyset$, if $i\neq j$? so that we can conclude $|B_1|+…+|B_k|=|B|=n$.
Best Answer
(1) is correct. In (2), the notation $\cup$ is troublesome. Let us view basis as indexed families rather than sets. Denote $B_i=(v_{i,1},\dots,v_{i,n_i})$ and $B=(v_{1,1},\dots,v_{1,n_1},\dots,v_{k,1},\dots,v_{k,n_k})$ (thus allowing a priori repetitions but we shall soon see there are none). Since $n_1+\dots+n_k=\dim V$ and $\operatorname{span}(B)=V,$ $B$ is a basis of $V.$