V is the image of T Let $T$ be a linear transformation from $U$ to $V$ then can we say that $V$ is always the image of $T$? what about when the $\ker T=\{0\}$?
Thanks for your helping.
 A: No. Even when $ker(T)=0$, it isn't true. For instance take $U$ to be any proper subspace of $V$, define $T: U \to V$ by $$T(u) = u.$$ The image of $T$ is $U$, which is not all of $V$.
A: No we can only say that $$Im(T)\subseteq V$$
Even when $ker(T)=0$ it isn't true.
Think for example to a matrix m-by-n: if $m>n$ you can have $ker(T)=0 $ (i.e. linearly indepedent columns) but the column can't span $\mathbb{R^m}$.
On the other hand, think also to a matrix m-by-n with $m<n$ in this case $ker(T)=0 $ (i.e. linearly depedent columns) but the column coul contain a basis and thus they span $\mathbb{R^m}$. In this case $Im(T)=V$.
A: The map $T\colon \mathbb{R} \rightarrow \mathbb{R}$ such that $T(x) = 0$ for any $x \in \mathbb{R}$ is linear, but $\text{Im}T = \{0\}$. 
When $\ker T = \{0\}$ by rank-nullity theorem you get 
$$\dim U  = \dim \text{Im} T.$$ 
A: Consider $U =\Bbb{R}^2, V = \Bbb{R}^3  $, $T:U\rightarrow V, (x_1,x_2) \mapsto (x_1,x_2,0)$. $T$ is linear, but $\text{im}(T) = \Bbb{R}^2 \times\{0\} \subsetneq V$ and $\text{ker}(T) = \{0\}$. (This is a specializing case of Ken Duna's Post)
