Prove that there exists a non-zero vector $v \in V$ such that $v \perp U$ 
$\DeclareMathOperator{\dim}{dim}$In Euclidean space $(X,\left\langle \cdot , \cdot \right\rangle )$ where $\dim X < \infty$. Suppose we have two subspaces $U,V \subset X$ and $\dim U<\dim V$. Prove that there exists a non-zero vector $v \in V$ such that $v \perp U$.

I tried to do this task however I don't have any idea how to prove it because I have less information about $U$ and I cannot even use the condition for the perpendicularity of vectors because it does not give me anything.
Can you help me?
 A: Let $p$ be the orthogonal projection on $U$, since $dimU <dim V$,  the kernel of the restriction of $p$ to $V$ is not zero ($dim V=dim ker p_{\mid V}+dim Imp_{\mid V}$ since $dim Imp_{\mid V}<dim U$  this implies that $dim kerp_{\mid V}>0$ since $dimU<dim V$), there exists $v\in V$ such that $p(v)=0$, this implies that $v$ is orthogonal to $U$.
A: We have that $U^\perp$ is a subspace of $X$ whose codimension is $\dim(U)$, while the codimension of $V$ is $\dim(X) - \dim(V)$.  Since the sum of these two codimensions is $\dim(X) - \dim(V) + \dim(U) < \dim(X)$, it follows that $U^\perp \cap V$ is not the trivial subspace.  Therefore, if you take a nonzero member $v \in U^\perp \cap V$, then $v \in V$ and $v$ is orthogonal to $U$.
A: A somewhat more concrete way to see this is as follows:
since
$\dim X < \infty, \tag 1$
we have
$\dim U < \infty \tag 2$
as well; let
$\vec e_i \in U, \; 1 \le i \le \dim U, \tag 3$
be an orthonormal basis (for $U$, of course).  For $\vec x \in X$ define
$P\vec x = \displaystyle \sum_1^{\dim U} \langle \vec x, \vec e_i \rangle \vec e_i; \tag 4$
it is easy to see that
$P:X \to U \tag 5$
is a linear map; also, for 
$\vec y \in U, \tag 6$
we have
$\vec y = \displaystyle \sum_1^{\dim U} \langle \vec y, \vec e_i \rangle \vec e_i, \tag 7$
thus, since $\langle \vec e_i, \vec e_j \rangle = \delta_{ij}$,
$P\vec y = \displaystyle \sum_{i = 1}^{\dim U} \left \langle \sum_{j = 1}^{\dim U} \langle \vec y, \vec e_j \rangle \vec e_j, \vec e_i \right \rangle \vec e_i = \sum_1^{\dim U} \langle \vec y, \vec e_i \rangle \vec e_i = \vec y; \tag 8$
that is, $P$ fixes $U$, element-wise.  Thus, in light of (5),
$\vec x \in X \Longrightarrow P^2 \vec x = P(P \vec x) = P \vec x, \tag 9$
i.e., 
$P^2 = P, \tag{10}$
a projection operator on $X$, with range $U$; a projection onto $U$.  Furthermore, it is also easy to see that $P$ is self-adjoint, that is,
$\vec x, \vec y \in X \Longrightarrow \langle P \vec x, \vec y \rangle = \langle \vec x, P \vec y \rangle; \tag{11}$
indeed,
$\langle P \vec x, \vec y \rangle = \left \langle \displaystyle \sum_1^{\dim U} \langle \vec x, \vec e_j \rangle  \vec e_j, \vec y \right \rangle =  \displaystyle \sum_1^{\dim U} \langle \vec x, \vec e_j \rangle \langle \vec e_j, \vec y \rangle = \sum_1^{\dim U} \langle \vec e_j, \vec y \rangle \langle \vec x, \vec e_j \rangle$
$= \displaystyle \sum_1^{\dim U} \langle \vec y, \vec e_j \rangle \langle \vec e_j, \vec x \rangle = \left \langle \displaystyle \sum_1^{\dim U} \langle \vec y, \vec e_j \rangle  \vec e_j, \vec x \right \rangle = \langle P\vec y, \vec x \rangle = \langle \vec x, P\vec y \rangle; \tag{12}$
now with (10) and (11) in hand we may resolve the primary question as follows:  since $\dim U < \dim V$ we may choose
$\vec v \in V \setminus U; \tag{13}$
then since
$\vec v \notin U, \tag{14}$
it follows that
$\vec v \ne P \vec v \in U; \tag{15}$
thus
$\vec v - P\vec v \ne 0; \tag{16}$
also, 
$\vec u \in U \Longrightarrow \langle \vec u, \vec v - P\vec v \rangle = \langle \vec u, \vec v \rangle - \langle \vec u, P \vec v \rangle = \langle \vec u, \vec v \rangle - \langle P \vec u,\vec v \rangle = \langle \vec u, \vec v \rangle - \langle \vec u, \vec v \rangle = 0, \tag{17}$
and we conclude that
$0 \ne \vec v - P \vec v \in U^\bot \tag{18}$
as required.
