# There are infinitely many subspaces to form a direct sum for $V$

Let $V$ be a finite dimensional vector space over $\mathbb R$ and let $U$ be a non-trivial proper subspace. Prove that there are infinitely many different subspaces $W$ of $V$ such that $V=U \oplus W$.

[Hint: Think first what happens when $V$ is $2$-dimensional; then generalise.]

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Have you thought about the hint at all...? – Zhen Lin Oct 25 '11 at 19:43
I am clear as to what happens when V is 2 dimensional, but I can't generalise from there. – WTY Oct 25 '11 at 19:55

So, in the 2-dimensional case, we have $U$, with basis $\mathbf{u}$ for some $\mathbf{u}_1\neq\mathbf{0}$. If $W$ is a subspace with $V=U\oplus W$, then letting $\mathbf{w}$ be a nonzero vector of $W$, we have that for every $\lambda\in\mathbb{R}$,

(i) $\lambda\mathbf{u}+\mathbf{w}\neq\mathbf{0}$;

(ii) $\mathrm{span}(\lambda\mathbf{u}+\mathbf{w})\cap U = \{\mathbf{0}\}$; and

(iii) $V = U + \mathrm{span}(\lambda\mathbf{u}+\mathbf{w})$.

(iv) $\mathrm{span}(\lambda\mathbf{u} + \mathbf{w}) = \mathrm{span}(\mu\mathbf{u}+\mathbf{w})$ if and only if $\lambda=\mu$.

To see (iii), note that $w\in U + \mathrm{span}(\lambda\mathbf{u}+\mathbf{w})$, hence $V= U+W = U+\mathrm{span}(\mathbf{w}) \subseteq U+\mathrm{span}(\lambda\mathbf{u}+\mathbf{w})$, so (iii) holds.

To verify (i) holds, note that if $\lambda\mathbf{u}+\mathbf{w}=\mathbf{0}$, then $\mathbf{w}= -\lambda\mathbf{u}\in U$, contradicting that $U\oplus W = V$.

For (ii), assume that $\mathbf{z}\in U\cap\mathrm{span}(\lambda\mathbf{u}+\mathbf{w})$. Then there exists $\kappa\in\mathbb{R}$ and $\mathbf{u}'\in U$ such that $\mathbf{z} = \mathbf{u}' = \kappa(\lambda\mathbf{u}+\mathbf{w}) = (\kappa\lambda)\mathbf{u} + \kappa\mathbf{w}$.

But from that equality we conclude that $\kappa\mathbf{w}=\mathbf{u}'-\kappa\lambda\mathbf{u}\in U$, so $\kappa\mathbf{w}=\mathbf{0}$, hence $\kappa=0$, so $\mathbf{z}=\mathbf{0}$.

Finally, to verify (iv), assume that the spans corresponding to $\lambda$ and to $\mu$ are equal. Then we can express $\mu\mathbf{u}+\mathbf{w}$ as a multiple of $\lambda\mathbf{u}+\mathbf{w}$. That is, there exists $\kappa\in\mathbb{R}$ such that $$\mu\mathbf{u}+\mathbf{w} = \kappa\lambda\mathbf{u}+\kappa\mathbf{w}.$$ This implies that $$(\mu - \kappa\lambda)\mathbf{u} = (\kappa-1)\mathbf{w}.$$ But then $(\kappa-1)\mathbf{w}\in W\cap U$, so $(\kappa-1)\mathbf{w}=\mathbf{0}$. This requries $\kappa-1=0$, so $\kappa=0$, and then $(\mu-\lambda)\mathbf{u} = \mathbf{0}$, so $\mu-\lambda=0$, hence $\mu=\lambda$.

So, for each real number $\lambda$, $W_{\lambda} = \mathrm{span}(\lambda\mathbf{u}+\mathbf{w})$ is a subspace such that $V=U\oplus W_{\lambda}$, and $W_{\lambda}=W_{\mu}$ if and only if $\lambda=\mu$. Since there are infinitely many choices for $\lambda$, there are infinitely many choices for $W$.

Can you generalize when expressed this way?

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So is this the correct way of generalising? Suppose U is a subspace as in question, with basis {u_1, ...u_k}. We take W st V=U⊕W, say W has a basis {w_1, ...w_(n-k)} We take Sp(λu_1+u_2+u_3+...+u_k+w_1+...+w_(n-k)) to be the subspace we need to complete the argument. – WTY Oct 25 '11 at 20:42
I think my previous comment works, after checking. Please correct me if I'm wrong. Anyway, thank you very much! Just for my learning purpose, how did you first decide to use Sp(λu+w)? I tried to use the hint by using lines in the Euclidean space, which obviously didn't help. – WTY Oct 25 '11 at 20:46
@WTY: Comment 2: If you start with, say, the horizontal and the vertical line in Euclidean space, then you can think of any line other than the vertical line as being determined by the vector $(1,0)+(0,m)$, where $m$ is the slope; this is $(1,0) + m(0,1)$, and so it goes from there. – Arturo Magidin Oct 25 '11 at 21:06
@WTY: Comment 1: That doesn't make a lot of esnse as written, because your W has dimension n-k, but the span you take is the span of a single vector. Perhaps you meant something else instead of a big sum of vectors? But I think you are making your life needlessly complicated by using all the vectors from $U$. Would it be enough to just "tweak" one of the basis vectors of $W$ using one of the basis vectors of $V$? – Arturo Magidin Oct 25 '11 at 21:07
Oops sorry, the "+" signs are supposed to be "," in comment 1. Thank you! – WTY Oct 27 '11 at 10:55