Annihilator of subspace in terms of a set difference?

Question

I am working through Linear Algebra Done Right by Axler, and I reached the chapter where Null(T'), the Null space of the dual of a linear map, is explored. I attempted to find the answer myself, and came up with this:

For T, a linear map from V to W:

$$Null(T') = (W - Range(T) \cup \{0\})'$$

where - denotes the set difference.

Once I read through the section, I realized Axler presents a theorem expressing Null(T') in terms of the annihilator $U^0$ of subspace U of a vector space V. The annihilator of U is defined to be the set of all linear functionals on V which map all elements of U to 0.

With this in mind, Axler proves that $Null(T') = (Range(T))^0$

As far as I can tell, this is equivalent to the expression I came up with. Can anyone verify this for me, or explain where I went wrong? I am a little new to the world of proofs, and I had trouble finding this question elsewhere online.

Answer 1

After considering Theo Bendit's comment, I have realized that my expression does not actually define a valid vector space, and that I was trying to describe a concept such as the orthogonal complement when I was writing it.

Answer 2

Kevin, welcome to math.stackexchange.com.

The statement in the original question that my book defines $$ \text{null } T' = (\text{range } T)^0 $$ is not correct. Instead, $\text{null } T'$ is defined (as usual) to be the subset of its domain that $T'$ sends to $0$. The equation above is then a theorem, not a definition, in the book (it is Theorem 3.107 in the third edition).