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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.

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  • $\begingroup$ Does $T'$ refer to the adjoint of $T$? I know you've referred to it as the "dual map", but I don't know of any such concept in Axler (though it's been a while since I've read it). Also, what does the ' mean when applied to sets? Is it the perpendicular complement, that is, does $S'$ mean the same as $S^\perp$? $\endgroup$ – Theo Bendit Sep 20 '18 at 23:17
  • $\begingroup$ In the 3rd edition, p. 103, T' is the notation for the 'dual map'. I am not yet familiar with the concept of adjoint (or I forgot since my first exposure to the subject!) $\endgroup$ – Kevin Bradner Sep 20 '18 at 23:19
  • $\begingroup$ The Dual Map T' of a linear map T is defined from the dual space of T's codomain to the dual space of T's domain, an it simply composes any linear functional on T's codomain with T $\endgroup$ – Kevin Bradner Sep 20 '18 at 23:21
  • $\begingroup$ Finally, the ' applied to sets only has meaning to me when that set is a vector space. As far as I am aware, the expression I found is a valid vector space. Then the ' simply is the space's dual, the set of all linear functionals on that space. $\endgroup$ – Kevin Bradner Sep 20 '18 at 23:22
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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.

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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).

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  • $\begingroup$ I have edited my question to reflect this. Please let me know if this has not addressed your concern correctly. Thanks for being a part of the community! It's cool to see you monitoring these questions. $\endgroup$ – Kevin Bradner Sep 21 '18 at 18:55
  • $\begingroup$ Hi Kevin, It is still not correct. To fix it, change "Axler defines" to "Axler proves that". $\endgroup$ – Sheldon Axler Sep 21 '18 at 22:10
  • $\begingroup$ got it, thank you. $\endgroup$ – Kevin Bradner Sep 21 '18 at 22:16

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