Give an example of a linear map $T$ such that dim [null ($T$)] = 3 and dim [range ($T$)] = 2. Give an example of a linear map $T$ such that dim [null ($T$)] = 3 and dim [range ($T$)] = 2. 
attempt: Suppose $V$ is finite dimensional. And let $T \in L(V,W)$.
The range $T$ is finite dimensional and dim $T$ = dim [null ($T$)]+[range ($T$)].
Then let $u_1,u_2,u_3$ be a basis of of null($T$); thus  dim [null ($T$)] = 3.
Then the linearly independent list $u_1,u_2,u_3$ can be extended to a basis $u_1,u_2,u_3,v_1,v_2$  of $V$. Thus dim $T$ = $3 + 2$ = $5$.
To see this, we will show that range of $T$ is finite dimensional and dim [range ($T$)] = $2$. We will show that $T{v_1}, Tv_2$ is a basis for range of $T$.
First , by definition of null we have $Tu_1 = Tu_2 = Tu_3 = 0$, and so dim [null ($T$)] = 3 .
Can someone please help me? Thanks.
 A: Your reasoning seems good, but you have not written the description of $T$. Therefore we can't say if your answer is right or wrong.
But let's go through what you have written. You have talked about a basis of $V $, namely $u_1,u_2,u_3,v_1,v_2$, out of which $u_i$ are in the null space. This directly translates to $T(u_i)=0$.
Now, all you need to do is ensure that $T(v_1)$ and $T(v_2)$ are linearly independent. The best way to do this is to simple consider two linearly independent vectors in $W$, say $w_1$  and $w_2$ (if $\dim W < 2$, you can't construct such a $T$!), and map $v_1 \to w_1$ and $v_2 \to w_2$.
Therefore, your full definition of the operator $T$ is like this: Let $x \in V$ be of the form $x = x_1u_1 + x_2u_2 + x_3u_3 + x_4v_1+x_5v_2$. Then, define $T: V \to W$ by $T(x) = x_4w_1 + x_5w_2$.
I leave you to check that $\dim [\text{null }T] = 3$ and $\dim [\text{range }T] = 2$.  
Just a friendly reminder : If $\dim W < 2$, then no such $T$ exists!

To give an explicit example : define $T : \mathbb R^5 \to \mathbb R^5$ by $T(a,b,c,d,e) = (a,b,0,0,0)$. This has rank two, nullity three. 
