I'm currently studying a book about Groups and Vector spaces and I read this in the section about direct sums of subspaces of Vector Spaces
Let $U$ be a subspace of $V$ and let $v_1,…,v_k$ be a basis of $U$. Extend $v_1,…,v_k$ to a basis $v_1,…,v_n$ of $V$ and let $w=sp(v_{k+1},…,v_n).$ Then $V=U \oplus W$. From this construction it follows there are infinitely many subspaces $W$ with $V = U \oplus W$ unless $U$ is $\{0\}$ or $V$
I can understand why if you extend $v_1,…,v_k$ to a basis $v_1,…,v_n$ of $V$ and let $w=sp(v_{k+1},…,v_n).$ Then $V=U \oplus W$, but I don't understand why there are guaranteed to be infinitely many ways of doing this. Wouldn't that depend on the vector space and field of scalars being used? Are there always infinitely many ways to extend a linearly independent set to a basis?
Best Answer
I don't think this is true in general.
Let $V := (\Bbb Z / 2\Bbb Z)^2$ be a vector space over $\Bbb Z / 2\Bbb Z = \{[0], [1]\}$. Let $U = \mathrm{span}\{e_1\} = \mathrm{span}\left\{ \begin{pmatrix} 1 \\0\end{pmatrix} \right\}$. Note that $V$ has only finitely many elements: $0,e_1, e_2, e_1+e_2$. From this it's easy to see that there are only finitely many choices for $W$ and the claim is false.
Some authors only consider real and complex vector spaces - in which case the claim would be true. Check the definition in your book.