Prove that $U^\perp$ = $\mathbb{R}^n$ if and only if U = {0}. Given that U is a subspace of $\mathbb{R}^n$, Show that $U^\perp$ = $\mathbb{R}^n$ if and only if U = {0}.
I am having difficulty solving this problem.  I can see that it makes sense, yet how would you prove this?  I feel like it can be solved with one of the vector projection theorems...
Likewise, can the opposite be proven? i.e.
Show that $U^\perp$ = {0} if and only if U = $\mathbb{R}^n$.
Thank you for any help!
 A: Let $\left\{u_{1},u_{2},\ldots ,u_{n}\right\}$ be a basis of $\mathbb{R}^{n}$. One can show that $$v = \sum_{i=1}^{n}{\frac{\left\langle v,u_{i}\right\rangle}{\left\langle u_{i},u_{i}\right\rangle} u_{i}}$$
Let $v \in U$, where $U^{\perp} = \mathbb{R}^{n}$. Then, in particular, $\left\langle v,u_{i}\right\rangle = 0$ for every $i = 1, \ldots, n$. By above result, $v = 0$; and hence, $U = \left\{0\right\}$.
OTOH, if $U = \{0\}$, it is easy to see that $U^{\perp} = \mathbb{R}^{n}$.
For the second question, observe that $\left(U^{\perp}\right)^{\perp} = U$. Then, if $U^{\perp} = \left\{0\right\}$, then $U = \left\{0\right\}^{\perp} = \mathbb{R}^{n}.$
A: We may try to directly tackle it.
Let $\langle \cdot, \cdot \rangle$ be an inner product on $\mathbb{R}^{n}$;
let $U$ be a vector subspace of $\mathbb{R}^{n}$. Then we have $U = \Bbb{R}^{n}$ only if $\langle x, 0 \rangle = 0$ for all $x \in U$, and only if $0 \in U^{\perp}$. If there is some $y \neq 0$ such that $y \in U^{\perp}$, then $\langle y,x \rangle = 0$ for all $x \in U$; but $\langle y, y \rangle \neq 0$ gives a contradiction.
Conversely, we have $U^{\perp} = \{ 0 \}$ only if $\langle x, 0 \rangle = 0$ for all $x \in U$. If $U \neq \mathbb{R}^{n}$, then there is some $y \in \Bbb{R}^{n}\setminus U$ such that $y \neq 0$ and $\langle y, 0 \rangle = 0$, implying that $y \in U^{\perp}$, a contradiction.
A: Since $U$ is a subspace of $\mathbb{R}^n$, we can always write: $$\mathbb{R}^n = U \oplus U^\perp. \tag{1}$$
Hence, it follows that $U^\perp = \{0\}$ if and only if $U=\mathbb{R}^n$.
