I'm having trouble with this problem. I know I should be using the theorem that $ \dim(\operatorname{im} T) = \dim(V) - \dim(\operatorname{Ker} T)$ to solve this question but I''m stuck on where to start.
For a linear map $T: \mathbb R^3 \rightarrow \mathbb R^2,$ what are the possible values of ($\dim \operatorname{Ker}T, \dim \operatorname{im} T$)?
The answer should be a pair of numbers. Thanks in advance.
$\def\ker{\operatorname{Ker}}\def\im{\operatorname{Im}}$ Since $\ker T$ is a subspace of $\mathbb R^3$, its dimension is $\leq3$; similarly $\dim\im T\leq2$. Further we have the constraint $\dim\ker T+\dim\im T=\dim\mathbb R^3=3$. So there are three possible pairs $(\dim\ker T,\dim\im T)=(3,0),(2,1),(1,2)$.
Here are examples of linear maps in each of the three cases:
$T:\mathbb R^3\rightarrow\mathbb R^2$ defined as
$ \begin{matrix} 1.&T(x,y,z)=(0,0),&\forall x,y,z\in\mathbb R.\\ 2.&T(x,y,z)=(x,0),&\forall x,y,z\in\mathbb R.\\ 3.&T(x,y,z)=(x,y),&\forall x,y,z\in\mathbb R.\\ \end{matrix} $
Hope this helps.