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.]
 A: 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?
