What are the possible dimensions of the kernel of T? I have the following question:
Let T be a linear map from $\Bbb{R}^5$ to $\Bbb{R}^3$. What are the possible dimensions of the kernel of T? Justify your answer.
I know by the Rank Nullity Theorem that $rankT$ and $nullT$ cannot be bigger than $dimV$, which in this case is 5, and the sum of the two should equal 5, however I'm not sure how to deduce the possible values for $rankT$ which I assume will then help me to deduce the possible values for $nullT$. 
Any suggestions would be helpful, thank you!
 A: Here’s one suggestion (not rigorous but to give you some intuition):
The kernel is the set of vectors in the domain that are mapped to zero in the codomain. The dimension of the kernel can be thought of as the number of dimensions that get ‘squashed’ by the transformation. By ‘squashed’, I mean, for example, all of the vectors in a $3$-dimensional space being mapped to a $2$-dimensional plane. You can imagine a cube, or some other $3$-dimensional object, being squashed until it is flat.
We are mapping from a $5$-dimensional space to a $3$-dimensional space, so we are already forced to squash $2$ dimensions. Therefore the dimension of the kernel is at least $2$. If all of the vectors are mapped to zero by the transformation, then all $5$ dimensions of the domain will be squashed, meaning that the dimension of the kernel is at most $5$. So we have $2 \leq dim(Null(T)) \leq 5$.
If you want to use the rank-nullity theorem, we can instead consider the image of $T$. In the case where all vectors are mapped to zero, the image clearly has dimension zero. It is also clear that the dimension of the image can be at most $3$, which will be the case if the ‘output’ vectors occupy all of the space we are mapping to. So we have $0 \leq dim(Im(T)) \leq 3$ which, by the rank-nullity theorem ($dim(Im(T)) + dim(Null(T)) = 5$ in this case), implies the result above.
A: By definition, if $V$ and $W$ are vector spaces and $T\colon V\to W$ is a linear transformation, then 
$$\text{Null}(T)=\{x\in V\,\colon T(x)=0\}.$$
Therefore, the nullspace of a linear transformation is a subset of its domain. In your case, $\text{Null}(T)\subseteq\mathbb{R}^5$ and hence $\text{dim}(\text{Null}(T))\leq 5$. On the other hand, $\text{Null}(T)$ contains at least one point, which is $0$ and thus $\text{dim}(\text{Null}(T))\geq 0$. Therefore, we concluded that you have $0\leq \text{dim}(\text{Null}(T))\leq 5$. Using the same reasoning, we conclude the more general statement that $0\leq \text{dim}(\text{Null}(T))\leq \text{dim}(V)$.
EDIT: As noted in the comments above, we can refine what I said above by using the Rank-Nullity Theorem. Using this result, we obtain that $\text{dim}(\text{Null}(T))= 5-\text{Rank}(T)$. Since the possible values for $\text{Rank}(T)$ are $0,1,2,$ and $3$, we obtain that $2\leq\text{dim}(\text{Null}(T))\leq 5$.
