A linear transformation whose domain has higher dimension than the target space I have a doubt in this question. I would like to check if my answer for letter a is correct and a hint for letter b.
Let $T:U\rightarrow V$ be a linear transformation where $U$ and $V$ are vector spaces such that $\dim_\mathbb{K}V<\dim_\mathbb{K}U < \infty$.
a) Prove the existence of a nonzero element $u \in U$ such that $T(u) = 0$.
b) Let $\mathbb{B}$ be an arbitrary basis of $U$. Does there always exists a vector $u \in \mathbb{B}$ such that $T(u) = 0$? Prove or give a counterexample.
My attempt:
a) Let $\dim_\mathbb{K}V = n$, $\dim_\mathbb{K}U = m$ and $B_U= \{ u_1, ..., u_n\}$ be a basis of U, so
$u$ = $\sum_{i=1}^m \alpha_i u_i \Longrightarrow T(u)  =  \sum_{i=1}^m \alpha_i T(u_i)$
$n = \dim_\mathbb{K}V < dim_\mathbb{K}U = m$, so all lineary independent sets in $V$ have at most 'n' elements, but
$A = \{ T(u_1), ..., T(u_m) \}$ has #$A = m > n$, therefore $A$ is lineary dependent. In other words,
$\sum_{i=1}^m \alpha_i T(u_i) = 0$ for some $0<i<m+1$. So, there exists a nonzero $u \in U$ such that $T(u) = 0$.
b)
For this question I have no  idea, but I think I need to use my development in 'a'. 
 A: We are given that $T:U\to V$ is a linear map where $U$ and $V$ are finite-dimensional vector spaces such that $\dim(V)<\dim(U)$. 
Here's a more efficient solution to (a) using the rank-nullity theorem.
Note that the rank-nullity theorem implies that
$$
\dim\bigl(\ker T\bigr)=\dim(U)-\dim\DeclareMathOperator{im}{im}\bigl(\im T\bigr)\tag{1}
$$
But $\im(T)$ is a subspace of $V$ so 
$$
\dim\bigl(\im T\bigr)\leq\dim V\tag{2}
$$
Combining (1) and (2) gives
$$
\dim\bigl(\ker T\bigr)\geq\dim(U)-\dim(V)\tag{3}
$$
But the inequality $\dim(V)<\dim(U)$ implies that $\dim(U)-\dim(V)>0$. Thus (3) gives
$$
\dim\bigl(\ker T\bigr)>0
$$
In particular, this implies that there exists a nonzero $u\in U$ such that $T(u)=\mathbf 0$. 
To address (b) let's write down some relevant examples. Linear maps $T:\Bbb R^2\to\Bbb R$ are always of the form $T(x,y)=ax+by$ for some $a,b\in\Bbb R$. We also have the standard basis $\{e_1=(1,0),e_2=(0,1)\}$ of $\Bbb R^2$. Can you find $a$ and $b$ such that $T(e_1)\neq0$ and $T(e_2)\neq 0$?
A: To correct your (a). Any vector in $U$ can be written as $\sum \alpha_i u_i$ and the corresponding image under $T$ can then be written as $\sum \alpha_i T(u_i)$. As you noted correctly $\{ T(u_1), \dots T(u_n) \}$ must be linearly dependent. By definition, this implies the existence of some scalars $\beta_1, \dots , \beta_n$, not all equal to $0$, such that $\sum \beta_i T(u_i) = 0$. And since $T$ is linear this gives $T(\sum \beta_i u_i) = 0$, i.e. if you define $u := \sum \beta_i u_i$, then $T(u) = 0$ and $u \neq 0$, since some $\beta_j$ is nonzero.
Another way to prove it, similar to the answer of @Brian Fitzpatrick, is to go by contradiction. Assume that no such $u$ exists. This implies that $T$ is injective, i.e. $V$ would contain a subspace, namely $T(U)$, which has higher dimension than $V$. 
Question (b) has been answered in the comments.
