When introducing dual spaces for the first time, most linear algebra textbooks proceed in what seems to me a rather backwards fashion: the annihilator $\{f\in V^*: f(u)=0\quad \forall u\in U\}$ of a subset/space $U$ of a vector space $V$ is introduced before the dual concept of the "joint kernel"(?) $\{v\in V: f(v)=0\quad \forall f\in W\}$ of a subset/space $W$ of the dual space $V^*$. The latter notion, despite corresponding to the intuitive idea of the solution space of a homogeneous linear system of equations, is then introduced indirectly by mapping $V$ into $V^{**}$ and considering the annihilator of the system of equations $W$ of $V^*$, which is $\{\phi\in V^{**}: \phi(f)=0\quad \forall f\in W\}$. This seems a pity as many students find the isomorphism of a finite dimensional vector space with its double dual difficult to grasp!

Worse, as the "(?)" above suggests, there appears to be no common terminology or notation for the concept dual to annihilator. Possibilities include "(joint) kernel, null/zero space, pre-annihilator, solution space". I'd be grateful for any pointers to textbooks which introduce the concept directly, or any suggestions for terminology and notation.

Update: Many thanks for the suggestions so far. I am still rather surprised and disappointed by the lack of references to elementary linear algebra text books which discuss solution spaces (aka null spaces, joint kernels) before or on the same footing as annihilators. I am unconvinced by the arguments that have so far been made for this omission.

Instead it seems to me that the current situation is the result of inertia. The "annihilator" is a term that caught on, and has been carried forward in the absence of a cool name for the dual concept. In this respect I like the invention of the "vanquished".

The question is not moot as I am currently lecturing this material. I am going to stick to "solution/null space" and "joint kernel". I will then discuss duality and row rank = column rank without ever mentioning the double dual (the double dual will come later in the course).

In preparing this, I noticed the fact that solution spaces and annihilators provide a Galois connection between the posets of subsets/subspaces of a vector space and its dual (that is, solution spaces and annihilators are contravariant adjoint functors between these posets). I've seen discussion of this in the context of annihilators in ring theory, but it seems to me to be at the heart of the concept of duality for vector spaces. I hope this rings some bells among MO readers.

Such categorical thinking leads to a fairly clean analysis of solution spaces and annihilators, which has in turn simplified some results subsequently.

  • $\begingroup$ Can you define your terminology a bit more clearly? Since there appears to be no common terminology or notation, perhaps you can illustrate it by an example, so others can be sure what concept you are talking about? $\endgroup$ Oct 16, 2010 at 20:27
  • 1
    $\begingroup$ I've added a couple of explicit definitions which I hope clarify what I am talking about. The dual space of a vector space V over a field F is the space of linear maps from V to F. $\endgroup$
    – David MJC
    Oct 16, 2010 at 20:36
  • 1
    $\begingroup$ I put excerpts from Nering below, I like the $\bot$ notation. He does not give an English phrase either. Given $W \subseteq V^*,$ I nominate "things smushed by $W$." $\endgroup$
    – Will Jagy
    Oct 16, 2010 at 21:47
  • 3
    $\begingroup$ How about "the vanquished of $W.$" $\endgroup$
    – Will Jagy
    Oct 16, 2010 at 22:04
  • 1
    $\begingroup$ I vaguely remember that when I was an undergraduate, we used orthogonal for the annihilator, and ante-orthogonal for your object (what comes "before" the orthogonal? orthogonal in the "previous" space?). It involved using the orthogonal notation $\perp$ for the orthogonal and an upside down version for the ante-orthogonal. I can see the gain in rigor offered by this, while hinting strongly at the inner product situation (that we had not yet covered). Still, I'm not sure how enlightening it was, and it did take me a bit before I realized the link with systems of equations. $\endgroup$ Oct 24, 2010 at 23:09

2 Answers 2


In the book "Algebra Lineal y Geometría" by Angel Rafael Larotonda, he introduces the concept right after introducing the annihilator of a set $M\in V$ as $M^o$, he introduces the "left" annihilator of a set $F \in V^*$ as $^oF=${$x \in V: f(x)=0 \; \forall \; f \in F$}, the only problem is that its written in spanish. I have also seen the concept introduced directly as pre-annihilator, but alas, also in spanish books.

  • $\begingroup$ This is the best answer so far (and I voted it up). However, the left annihilator still has a secondary role ("pre-annihilator" makes the secondary nature of the concept even more plain) despite the fact that "solution space" ("vectors satisfying the equations") is a more fundamental and intuitive concept than "annihilator" ("equations which the vectors satisfy"). Are all our linear algebra books this bad with regard to dual spaces? Or is there a textbook which explains that the dual space is the space of equations, and that subspaces of equations have solution spaces? $\endgroup$
    – David MJC
    Oct 16, 2010 at 23:50
  • 2
    $\begingroup$ In my opinion, it isn't that "all our linear algebra books are this bad". It's the mathematic trend to teach the concept from which you can branch a lot of stuff than the motivation, concrete examples. I think this mathoverflow.net/questions/19152/… question in MO can explain very clearly what I have in mind. $\endgroup$ Oct 17, 2010 at 0:38
  • $\begingroup$ Those are noble goals, but are they really why annihilators are given primacy ahead of null spaces? After all, we don't usually define continuity in metric spaces first in terms of open sets, then give the metric definition. Also, unlike open sets, annihilators do not yield a whole new generalized foundation for linear algebra (that I know of!). They are just part of the toolkit, and work best in the finite dimensional case, where they are an equivalent dual notion to null spaces. In general, annihilators have the advantage over null spaces that they iterate - perhaps that is what you mean? $\endgroup$
    – David MJC
    Oct 18, 2010 at 20:16

Let $V_0\leq V$ be a subspace of the vector space $V$. If $ann(V_0)\leq V^*$ is its annihilator, then the dual of $ann(V_0)$ is isomorphic with the quotient space $V/V_0$:

$ann(V_0) \cong \frac{V}{V_0}$

This link may be useful: http://en.wikipedia.org/wiki/Dual_space#Quotient_spaces_and_annihilators


Oops, after I added my answer, the page refreshed and I saw that you edited your comment. It seems that my answer does not apply to your question...

Anyway, I see no reason to give it a different name than annihilator, even if it annihilates a subspace of $V^*$. It has the same definition.

Update 2

It may depend on the context:

In functional analysis some use the notion of pre-annihilator because it is important to distinguish $V^{**}$ from $V$. Example: Dales, Aiena - Introduction to Banach Algebras, Operators and Harmonic Analysis.

When $V^{**}\cong V$ some don't necessarily make the distinction. Example: Marsden and Ratiu - The Breadth of Symplectic and Poisson Geometry

But it may be a good practice to specify in what space a subspace is the annihilator of a given subspace: Example: In Automorphic Forms on GL(2), Jacquet and Langlands said: "If $\tilde V_2$ is the annihilator of $V_1$ in $\tilde V$ then $V_1$ is the annihilator of $\tilde V_2$ in $V$.

For non-commutative rings, ideals, semigroups, one specifies whether the annihilator is left or right. Examples: G. Gratzer - Universal Algebra Steven G. Krantz (Ed.) - Dictionary of Algebra, Arithmetic, and Trigonometry

  • $\begingroup$ Apologies for my confusing initial post: by "dual concept", I did not mean "dual space". In the notation of my post, the annihilator of $W$ is a subspace of $V^{**}$, not a subspace of $V$, so the definition is not the same. $\endgroup$
    – David MJC
    Oct 16, 2010 at 21:21
  • $\begingroup$ After reading your updated question I find that at least its title was clearly formulated from the beginning. So it's my mistake :) $\endgroup$ Oct 16, 2010 at 21:57
  • $\begingroup$ Many thanks for the references. I have also seen authors making the "annihilator in..." distinction. $\endgroup$
    – David MJC
    Oct 18, 2010 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.