If the dimension of the domain of a linear transformation is smaller than that of the codomain, the transformation is not surjective always? Given two vector spaces $V$ and $W$, defined over the same field $K$, and a linear transformation between them $T:V\longrightarrow W$, such that $\text{dim}(V) = n$, $\text{dim}(W) = m$ and $n \lt m$, there exists a vector $w \in W$ such that $w \notin \text{Im}(T)$ (i.e. $T$ is not surjective)?
 A: The dimension theorem states that $$\dim(V)=\dim(\ker(T))+\dim(\text{im}(T)).$$
The dimension of $\text{im}(T)$ is maximal if $T$ is injective ($\dim(\ker(T))=0$). But then $n=\dim(\text{im}(T))<m$, hence $T$ is not surjective.
Edit: Let's include a proof of the dimension theorem for linear maps. 
Let $\left\{v_1,\dots ,v_k\right\}$ be a basis for $\ker(T)$. Extend this basis to a basis for all of $V$: $\left\{v_1,\dots , v_k,w_{k+1},\dots ,w_n\right\}$. Then $\text{Im}(T)=S[T(v_1),\dots, T(v_k),T(w_{k+1}),\dots, T(w_n)]=S[T(w_{k+1}),\dots ,T(w_n)]$ since $T(v_i)=0$. (Here $S[\dots]$ denotes the span of those vectors). If we can show that the $T(w_j)$ are independent, then $\dim(\text{im}(T))=n-k$ and hence $\dim(V)=\dim(\ker(T))+\dim(\text{im}(T))$. 
So suppose that $\sum_{j=k+1}^n \mu_jT(w_j)=0$, which is equivalent to $T(\sum_{j=k+1}^n \mu_jw_j)=0$. Now, if $\sum_{j=k+1}^n \mu_jw_j=0$, then every $\mu_j$ must be zero because $w_j$ are linearly independent, in which case the theorem would be proven. Alternatively, $\sum_{j=k+1}^n \mu_jw_j\in \ker(T)$. If this were the case, we can then write $\sum_{j=k+1}^n \mu_jw_j$ in the basis of $\ker(T)$. Thus $\sum_{j=k+1}^n \mu_jw_j=\sum_{i=1}^k\lambda_iv_i$, equivalently, $\sum_{i=1}^k\lambda_iv_i-\sum_{j=k+1}^n \mu_jw_j=0$. But then we have a linear combination of basis vectors equal to zero, hence $\lambda_i,\mu_j$ are all zeroes, again proving the theorem.
