How to see the Image, rank, null space and nullity of a linear transformation

Question

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?

Answer 1

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.

Answer 2

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.