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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?

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  • $\begingroup$ Indeed, this depends on the field of scalars: if it's a finite field, then there are only finitely many vectors. But this bolded claim is true for infinite fields. $\endgroup$
    – Dave
    Oct 23, 2021 at 14:02

3 Answers 3

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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.

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  • $\begingroup$ You're correct. After reading your answer I checked, and when Vector Spaces are first introduced in the book, it is specified that F is either $\mathbb{R}$ or $\mathbb{C}$ Thank you! $\endgroup$
    – 123123
    Oct 23, 2021 at 14:15
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Notice that for an intutive example, you can think of $ \mathbb{R}^2$, and U which is given by span of ${(1,0)}$. Then you can take either span$(1,1)$ or span$(0,1)$, and so on. The only condition here is the indepndence which yields cleraly infinite vectors that are indep of $(1,0)$.

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Yes, it depends upon the field of scalars. But if it is an infinite field, then it is true. For instance, if $u\in U$ and $\lambda$ is a scalar, consider$$W_\lambda=\operatorname{span}(\{v_{k+1}+\lambda u,v_{k+2},\ldots,v_n\}).$$Then $\lambda\ne\lambda'\implies W_\lambda\ne W_{\lambda'}$.

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