Proof: $B$ is a basis of $V$ $\Longleftrightarrow$ $V = \bigoplus W_k$

Question

Let $V$ be any (finite) vector space, $\oplus$ donate the (inner) direct sum, and $B_k$ is a given basis for some supspace $W_k\subset V$ and $B=\bigcup_k B_k$ then $$V = \bigoplus_k W_k \Longleftrightarrow B \text{ is a basis of } V$$

The statement is clear to me, but i have some problems to show it formaly.

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I got it up to:

$$\begin{eqnarray} V = \bigoplus_k W_k &\Longleftrightarrow& V = W_1 + \dots W_n \bigwedge w_1\in W_1\dots w_n\in W_n \text{ are linear independent } (w_i\neq0)\\ &\Longleftrightarrow& V = \sum_k\operatorname{span}\left(B_k\right) \bigwedge B_k \text{ is a basis for } W_k \\ && \bigwedge w_1\in W_1\dots w_n\in W_n \text{ are linear independent } (w_i\neq0)\\ &\Longleftrightarrow& \dots \\ &\Longleftrightarrow& V =\operatorname{span}\left(\bigcup B_k\right) \bigwedge \text{ the elements of }\left(\bigcup B_k\right)\text{ are linear independent}\\ &\Longleftrightarrow& V = \operatorname{span}(B) \bigwedge \text{ the elements of }B\text{ are linear independent}\\ &\Longleftrightarrow& B \text{ is a basis of } V \end{eqnarray} $$

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I do not know how to formulate the last 1-2 steps in the middle. Thanks in advance.

Answer 1

As said, this is wrong: take $V = \mathbb{R}^2$ and $W_1 = W_2 = V$ with basis $B_1 = B_2 = \{e_1,e_2\}$ (since you write a union for the basis, I suppose that you consider that a basis is a set and not an ordered family). Of course, $W_1$ and $W_2$ are not in direct sum but $B = B_1 \cup B_2 = B_1$ is a basis of $V$.

This should be rewritten by taking ordered families for the bases: $B_k = (f^k_1,\dots,f^k_{l(k)})$ and $B= (f^1_1,\dots,f^1_{l(1)},\dots,f^s_1,\dots,f^s_{l(s)})$ where $s$ is the number of $W_k$. Then:

\begin{align} V = \oplus_k W_k &\Longleftrightarrow \forall x \in V, \forall (1 \leq k \leq s), \exists! w_k \in W_k: x = \sum_{k=1}^s w_k \\ &\Longleftrightarrow \forall x \in V, \forall (1 \leq k \leq s), \forall (1 \leq i(k) \leq l(k)), \exists! \lambda^k_{i(k)}\in \mathbb{R}: x = \sum_{k=1}^s \sum_{i(k)=1}^{l(k)} \lambda^k_{i(k)} f^k_{i(k)} \\ & \Longleftrightarrow B \, \text{ is a basis of $V$} \end{align}