## Theorem Relating Direct Sums, Annihilators, and Dual Spaces

**Point of post:** In this post I give a proof of a theorem which, given gives the explicit relationship between and . I also give precursory proofs regarding the formation of bases for annihilators, dual spaces and direct sums, mostly for my own benefit. Any reader who knows these theorems is free to skip them and move directly to the seminal theorem. The culmination of these theorems will be that

*N.B.* This is all regarding finite dimensional vector spaces.

**Lemma:** Let be an -dimensional vector space over a field . Then, .

**Proof:** Let be a basis for . Clearly then, as was proven before there are unique such that . We claim that is a basis for . To see this first suppose that

Then, we see that

from where the fact that follows. Now, let be arbitrary. We claim that . To see this, let then for some and so

Thus, is a basis for and so as desired.

*Remark:* The only reason for proving the above, was so that one can see, in action the idea of the *dual basis.* Namely, given a basis the associated basis denoted above as is the dual basis associated with . In fact, a much more general result can be found here.

We next prove an analogous result for the annihilator

**Lemma: **Let be a -dimensional subspace of an -dimensional vector space . Then, is an -dimensional subspace of

**Proof: **Since is -dimensional it has a basis which may be extended to a basis for . Let be the associated dual basis, we claim that is a basis for . Clearly since (being a subset of a linearly independent set) is linearly independent it suffices to prove that . To see this let . Then, since is a basis for there exists such that

But, for we see that the left hand side is since and , and the right hand side equations by definition of the dual base. It follows that so that

from where immediately follows, and thus is a basis for of cardinality as required.

Lastly, we prove a theorem about the basis for a direct sum (here interpreted as the disjoint internal sum), from where upon inspection the theorem itself will follow immediately (though we will write an actual proof)

**Theorem: **Let and be and -dimensional subspaces of the -dimensional vector space and assume that . Then, if and are bases for and respectively (here , for notational convenience).

**Proof: **Suppose that

then

from where it follows from that

from where the linear independence of and respectively we see that

from where the linear independence of follows. Now, we know since that for every we have that for some and . But, there exists such that and and so

from where follows.

So, from here the intuition of the theorem now becomes clear, although there is some subtlety to the actual proof (not much). Namely, the intuition goes diagrammatically as follows: if and

where canonical just means it’s the basis that you would pick if you were to pick a basis based on the given info. And, the two-sided vertical arrow is meant to indicate the isomorphism we’re going to form. So, without further ado.

**Theorem:** Let and be subspaces of , with the property that . Then, , , and .

**Proof:** Define

To show that it’s linear we notice that evidently

Also, a quick (this is less obvious, but true) check shows that

Next, we note that if and are distinct elements of , then there exists such that . But, since for all we may conclude that . Thus, for some , and thus . To prove surjectivity we appeal to a dimension argument noticing that is a subspace of of dimension but, and so , from where the conclusion follows.

*Remark:* Really what we did above, could be in a completely non-categorical sense, considered an argument by abstract nonsense. We appealed to a theorem of dimension and really circumvented the issue at hand, mainly for convenience. Note though, the diagram argument was meant to show interested readers how to take a more direct route to prove surjectivity. Namely, the basis elements for are precisely the restrictions to of the basis elements of . So, for every basis element we may extend by defining and show that this is precisely one of the basis elements of where the rest of the argument follows from extension by linearity.

A completely analogous argument shows that .

So, it remains to prove that . To do this we first note that if then , but since the zero functional is the unique functional with this property we may conclude that . We, once again, appeal to a dimension argument to finish. Namely that since we see that

and so the conclusion follows.

*Remark:* While the same comment regarding the dimension argument may apply, it does so with less oomph. If then a quick check gives that the first are a basis for and the last a basis for , and with their trivial intersection this says that their sum (as subspaces) has the union of these two bases, which is precisely the basis for .

*Remark:* In fact, if one pays close attention the entirety of the above has been a farce, if one realizes that the spaces we are trying to show are isomorphic have the same dimension. That said, the methodology used here leaves interpretation for infinite dimensional space analogues.

**References:**

1. Halmos, Paul R. “Dual Bases, Reflexivity, Annihilators.” *Finite-dimensional Vector Spaces,*. New York: Springer-Verlag, 1974. 23-28. Print.

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