Direct sum proof: If the union of the basis of $W_1,W_2,...,W_k$ is a basis of $V$ than $W_1 \oplus W_2 \oplus ... \oplus W_k= V$

Question

Considering a finite dimensional vector space $V$ and $k$ of its subspaces $W$ is it possible to say that

If the union of the basis of $W_1,W_2,...,W_k$ is a basis of $V$ than $W_1 \oplus W_2 \oplus ... \oplus W_k= V$

?

Can anyone give a counterexample if it is not true?

Should I consider in the hypotesis that any basis of $W_1,W_2,...,W_k$ is a basis of $V$ ? Would it work in that case?

Thanks in advice

Answer 1

Neither version quite does the trick, but you're close.

What we can say is the following:

Suppose $W_1,...,W_n$ are subspaces of $V,$ and $\mathcal B_1,...,\mathcal B_n$ (respectively) are bases of $W_1,...,W_n$ that are pairwise disjoint. Then $\mathcal B_1\cup\cdots\cup\mathcal B_n$ is a basis of $V$ if and only if $W_1\oplus\cdots\oplus W_n=V.$

However, the converse holds. In particular, we have the following:

Suppose $W_1,...,W_n$ are subspaces of $V$ such that $W_1\oplus\cdots\oplus W_n=V.$ Then for any bases $\mathcal B_1,...,\mathcal B_n$ (respectively) of $W_1,...,W_n,$ we have that $\mathcal B_1\cup\cdots\cup\mathcal B_n$ is a basis of $V.$

Answer 2

Consider the space $R^3$, with the coordinate planes as subspaces, using the standard basis.The union of the bases for the three coordinate planes is the standard basis on $R^3$, but the dimension of the direct product is 6, clearly not $R^3$.