How to see the Image, rank, null space and nullity of a linear transformation 
If V is a vector space over the field F, in each case, find the Image, rank, null space and nullity of the given linear transformation :
a) $0$: V → V defined as $0(v)$ := $0$
b) $Id_V$ : V → V defined as $Id(v)$ := v
c)T : V → V defined as $T(x, y)$ := (x − y, 0).

a) Image: 0;
rank: 1;
null space: All v $\in$ V;
nullity: dim$_F(V)$
b) Image: v;
rank: dim$_F(V)$;
null space: $0$;
nullity: dim$_F(0)$=1
c) Image: (x-y,0);
rank: 2;
null space: (x,x), i.e (x,y) where x=y;
nullity: 1
I am almost sure I am getting these things wrong, because to verify that I can use the theorem that rank(T) + null(T) = dim$_F$(V).
What mistakes am I making?
 A: a) The image is $\{0\}$, not $0$. And the rank is $0(=\dim\{0\})$. You are right about the null space and the nullity.
b) The image is $V$, not $v$. And, although indeed the nullity is $\dim_F\{0\}$, I would have said that it is $0$.
c) The image if $F\times\{0\}$, and the rank is $1$. The rest is fine.
A: a) Im(0) = {0}. Remember that rank(T) = dim(Im($T$)), where $T$ is a linear transformation. But dim({0}) = 0 (see Wikipedia), hence Rank(0) = 0. The null space is Ker(0) = $V$. Nullity(0) = dim$_F$($V$).
b) Im(Id$_V$) = $V$ (the set $V$, not the element $v$). Rank(Id$_V$) = dim$_F$($V$). Ker(Id$_V$) = {0}. Nullity(Id$_V$) = 0.
c) May I interpret $T$ as $T: V \times V \to V \times V$. Then Im($T$) = $V \times 0$ (not $F$). But I gues $T(x, y)$ means $T(v)$, where $v = (x, y), v \in V$. So in this case we have Im$(T) = \{v: v = (x, 0), x \in F \} = F \times \{0\}$. And Rank$(T) = 1$. Ker$(T) = \{v: v = (x, x), x \in F \}$. Nullity$(T) = 1$. Note that dim$_F$($V$) = 2 and the Rank-nullity theorem applies.
