Linear transformations, a general question Let a linear transformation $R: U \rightarrow W$, $U, W$ are final-dimensional vector spaces. Let $(u_1, ...,u_n)$,  $(w_1, ... w_m)$ be basis of $U$, $W$ s.t $n > m$. Suppose that for each $ 1\leq j\leq m \space R(u_j) = w_j$. Why or is it necessary that for each $ m+1\leq j\leq n \space \space R(u_j) = 0$? I think the abstraction makes it harder for me to see how it relates to the kernel
 A: You might have heard that given a basis of $U$, a linear transformation $R : U\to W$ is fully specified by its action on the basis vectors of $U$.
Any (ordered) choice of $n=\dim U$ vectors $w_1,\ldots,w_n$ in $W$ uniquely determines a linear transformation with $R(u_j)=w_j$. And vice versa - given $R$ and a basis of $U$, you have a set of $n$ vectors $R(u_1),\ldots,R(u_n)$ that determines $R$.
In your case, you're given only the first $m$ vectors. As $m<n$ and considering the above, the remaining $(n-m)$ vectors required to specify $R$ are "free" to be chosen (from $U$) in any way you want - not necessarily as the zero vector.

Here's another way to look at it: since you have the bases for both spaces, it might help to write out the matrix of $R$.
The condition $R(u_j)=w_j$ for $1\le j \le m$ means exactly that the first $m$ columns of $R$'s matrix consist of $(0,\ldots,0,1,0,\ldots 0)$. For example, for $n=6,m=3$ the matrix looks like this:
$$
\begin{pmatrix}1 & 0 & 0 & w & h & a\\
0 & 1 & 0 & t & e & v\\
0 & 0 & 1 & e & r & :)
\end{pmatrix}
$$
Your condition fixed the lefmost "identity square", but did not say anything about the rest of the matrix. Hence the rest can be filled with any scalars you want, not necessarily zeros.
A: Suppose that $U$ has basis $u_1,u_2$ and that $W$ has basis $w_1$ and that $R$ is the linear transformation $R(au_1+bu_2)=(a+b)w_1$. Then $R(u_2)\ne 0$.
A: The rank-nullity theorem states that
$$\dim\operatorname{im} (T)+\dim\operatorname{Ker}(T)=\dim V$$
so
$$\dim W+\dim\operatorname{Ker}(R)=\dim V$$
$$\dim\operatorname{Ker}(R)=n-m$$
As $R$ transforms one basis into another, we must have the kernel of $R$ ($R(u_k)=0$ ) with dimension $n-m$, i.e. for $m+1\le k\le n$.
