# Abstract Nonsense

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

Point of post: In this post I give a proof of a theorem which, given $\mathscr{W}=\mathscr{U}\oplus\mathscr{V}$ gives the explicit relationship between $\text{Ann }\mathscr{V},\text{Ann }\mathscr{U},\text{Hom}\left(\mathscr{U},F\right),\text{Hom}\left(\mathscr{V},F\right)$ and $\text{Hom}\left(\mathscr{W},F\right)$.  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

$\mathscr{W}=\mathscr{U}\oplus\mathscr{V}\implies \text{Hom}\left(\mathscr{W},F\right)=\text{Ann }\mathscr{U}\oplus\text{Ann }\mathscr{V}$

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

Lemma: Let $\mathscr{V}$ be an $n$-dimensional vector space over a field $F$. Then, $\dim_F \text{Hom}\left(\mathscr{V},F\right)=n$.

Proof: Let $\{e_1,\cdots,e_n\}$ be a basis for $\mathscr{V}$. Clearly then, as was proven before there are unique $\varphi_j\in\text{Hom}\left(\mathscr{V},F\right),\text{ }j=1,\cdots,n$ such that $[e_i,\varphi_j]=\delta_{ij}$. We claim that $\{\varphi_1,\cdots,\varphi_n\}$ is a basis for $\text{Hom}\left(\mathscr{V},F\right)$. To see this first suppose that

$\displaystyle \sum_{j=1}^{n}\alpha_j\varphi_j=\bold{0}$

Then, we see that

$\displaystyle \sum_{j=1}^{n}\alpha_j\varphi_j(e_i)=\alpha_i,\text{ }i=1,\cdots,n$

from where the fact that $\{\varphi_1,\cdots,\varphi_n\}$ follows. Now, let $\varphi\in\text{Hom}\left(\mathscr{V},F\right)$ be arbitrary. We claim that $\displaystyle \varphi=\sum_{j=1}^{n}\varphi(e_j)\varphi_j$. To see this, let $v\in\mathcal{V}$ then $v=\sum_{k=1}^{n}\alpha_k x_k$ for some $\alpha_1,\cdots,\alpha_n$ and so

\displaystyle \begin{aligned} \sum_{j=1}^{n}\varphi(e_j)\varphi_j(v) &= \sum_{j=1}^{n}\varphi(e_j)\varphi_j\left(\sum_{k=1}^{n}\alpha_k e_k\right)\\ &= \sum_{j=1}^{n}\left(\varphi(e_j)\sum_{k=1}^{n}\alpha_k \varphi_{j}(e_k)\right)\\ &= \sum_{j=1}^{n}\alpha_j \varphi(e_j)\\ &=\varphi\left(\sum_{k=1}^{n}\alpha_k e_k\right)\\ &=\varphi(v) \end{aligned}

Thus, $\{\varphi_1,\cdots,\varphi_n\}$ is a basis for $\text{Hom}\left(\mathscr{V},F\right)$ and so $\dim_F\text{Hom}\left(\mathscr{V},F\right)=n$ as desired. $\blacksquare$

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 $\{e_1,\cdots,e_n\}$ the associated basis denoted above as $\{\varphi_1,\cdots,\varphi_n\}$ is the dual basis associated with $\{e_1,\cdots,e_n\}$. In fact, a much more general result can be found here.

We next prove an analogous result for the annihilator

Lemma: Let $\mathscr{M}$ be a $m$-dimensional subspace of an $n$-dimensional vector space $\mathscr{V}$. Then, $\text{Ann }\mathscr{M}$ is an $n-m$-dimensional subspace of $\text{Hom}\left(\mathscr{V},F\right)$

Proof: Since $\mathscr{M}$ is $m$-dimensional it has a basis $\{x_1,\cdots,x_m\}$ which may be extended to a basis $\{x_1,\cdots,x_m,\cdots,x_{m+1},\cdots,x_n\}$ for $\mathcal{N}$. Let $\{\varphi_1,\cdots,\varphi_m,\varphi_{m+1},\cdots\varphi_n\}$ be the associated dual basis, we claim that $\{\varphi_{m+1},\cdots,\varphi_n\}$ is a basis for $\text{Ann }\mathscr{M}$. Clearly since $\{\varphi_{m+1},\cdots,\varphi_{n}\}$ (being a subset of a linearly independent set) is linearly independent it suffices to prove that $\text{span }\{\varphi_{m+1},\cdots,\varphi_n\}$. To see this let $\varphi\in\text{Ann }\mathscr{M}$. Then, since $\{\varphi_1,\cdots,\varphi_{m},\varphi_{m+1},\cdots,\varphi_n\}$ is a basis for $\text{Hom}\left(\mathscr{V},F\right)$ there exists $\alpha_1,\cdots,\alpha_n$ such that

$\displaystyle \varphi=\sum_{j=1}^{n}\alpha_j \varphi_j$

But, for $x_k,\text{ }k\in\{1,\cdots,m\}$ we see that the left hand side  is $0$ since $x_k\in\mathscr{M}$ and $\varphi\in\text{Ann }\mathscr{M}$, and the right hand side equations $\alpha_k$ by definition of the dual base. It follows that $\alpha_1=\cdots=\alpha_m=0$ so that

$\displaystyle v=\sum_{j=m+1}^{n}\alpha_j\varphi_j$

from where $\text{span }\{\varphi_{m+1},\cdots,\varphi_m\}=\text{Ann }\mathscr{M}$ immediately follows, and thus $\{\varphi_{m+1},\cdots,\varphi_n\}$ is a basis for $\mathscr{M}$ of cardinality $n-m$ as required. $\blacksquare$

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 $\mathscr{U}$ and $\mathscr{V}$ be $m$ and $n-m$-dimensional subspaces of the $n$-dimensional vector space $\mathscr{W}$ and assume that $\mathscr{W}=\mathscr{U}\oplus\mathscr{V}$. Then, if $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_k\}$ are bases for $\mathscr{U}$ and $\mathscr{V}$ respectively (here $k=n-m$, for notational convenience).

Proof: Suppose that

$\displaystyle \sum_{j=1}^{n}\alpha_j x_j+\sum_{j=1}^{k}\beta_j y_j=\bold{0}$

then

$\displaystyle \sum_{j=1}^{n}\alpha_j x_j=\sum_{j=1}^{k}(-\beta_j)y_j$

from where it follows from $\mathscr{U}\cap\mathscr{N}=\{\bold{0}\}$ that

$\displaystyle \sum_{j=1}^{n}\alpha_j x_j=\sum_{j=1}^{k}\beta_j y_j=\bold{0}$

from where the linear independence of $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_k\}$ respectively we see that

$\alpha_1=\cdots=\alpha_m=\beta_1=\cdots=\beta_k=0$

from where the linear independence of $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ follows. Now, we know since $\mathscr{W}=\mathscr{U}\oplus\mathscr{V}$ that for every $w\in\mathscr{W}$ we  have that $w=u+v$ for some $u\in\mathscr{U}$ and $v\in\mathscr{V}$. But, there exists $\alpha_1,\cdots,\alpha_m,\beta_1,\cdots,\beta_k\in F$ such that $\displaystyle u=\sum_{j=1}^{m}\alpha_j x_j$ and $\displaystyle v=\sum_{j=1}^{n}\beta_j y_j$ and so

$\displaystyle w=u+v=\sum_{j=1}^{m}\alpha_j x_j+\sum_{j=1}^{k}\beta_j y_j$

from where $\mathscr{W}=\text{span }\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ follows. $\blacksquare$

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 $\{x_1,\cdots,x_m\}$ and $\{y_1,\cdots,y_k\}$



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 $\mathscr{U}$ and $\mathscr{V}$ be subspaces of $\mathscr{W}$, with the property that $\mathscr{W}=\mathscr{U}\oplus\mathscr{V}$. Then, $\text{Ann }\mathscr{V}\cong\text{Hom}\left(\mathscr{U},F\right)$, $\text{Ann }\mathscr{U}\cong\text{Hom}\left(\mathscr{V},F\right)$, and $\text{Hom}\left(\mathscr{W},F\right)=\text{Ann }\mathscr{U}\oplus\text{Ann }\mathscr{V}$.

Proof: Define

$f:\text{Ann }\mathscr{V}\to\text{Hom}\left(\mathscr{U},F\right):\varphi\mapsto\varphi_{\mid \mathscr{U}}$

To show that it’s linear we notice that evidently

$f(\alpha\varphi)=(\alpha\varphi)_{\mid \mathscr{U}}=\alpha\varphi_{\mid \mathscr{U}}=\alpha f(\varphi)$

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

$f\left(\varphi_1+\varphi_2\right)=\left(\varphi+\psi\right)_{\mid \mathscr{U}}=\varphi_{\mid \mathscr{U}}+\psi_{\mid\mathscr{U}}=f(\varphi)+f(\psi)$

Next, we note that if $\varphi$ and $\psi$ are distinct elements of $\text{Ann }\mathscr{V}$, then there exists $w\in\mathscr{W}$ such that $\varphi(w)\ne\psi(w)$. But, since $\varphi(w)=\psi(w)=0$ for all $w\in\mathscr{V}$ we may conclude that $w\in\mathscr{U}$. Thus, $\varphi(w)\ne\psi(w)$ for some $w\in\mathscr{U}$, and thus $f(\varphi)=\varphi_{\mid\mathscr{U}}\ne\psi_{\mid\mathscr{U}}=f(\psi)$. To prove surjectivity we appeal to a dimension argument noticing that $f\left(\text{Ann }\mathscr{V}\right)$ is a subspace of $\text{Hom}\left(\mathscr{U},F\right)$ of dimension $\dim_F\mathscr{W}-\dim_F\mathscr{V}=\dim_F\mathscr{U}$ but, $\mathscr{U}\cong\text{Hom}\left(\mathscr{U},F\right)$ and so $\dim_F f\left(\text{Ann }\mathscr{V}\right)=\dim_F\text{Hom}\left(\mathscr{U},F\right)$ , 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 $\text{Hom}\left(\mathscr{U},F\right)$ are precisely the restrictions to $\mathscr{U}$ of the basis elements of $\text{Ann }\mathscr{V}$. So, for every basis element $\varphi\in\text{Hom}\left(\mathscr{U},F\right)$ we may extend $\varphi:\mathscr{U}\to F$ by defining $[y_\ell,\varphi]=0,\text{ }\ell=1,\cdots,k$ and show that this is precisely one of the basis elements of $\text{Ann }\mathscr{V}$ where the rest of the argument follows from extension by linearity.

A completely analogous argument shows that $\text{Ann }\mathscr{U}\cong\text{Hom}\left(\mathscr{V},F\right)$.

So, it remains to prove that $\text{Hom}\left(\mathscr{W},F\right)=\text{Ann }\mathscr{U}\oplus\text{Ann }\mathscr{V}$. To do this we first note that if $\varphi\in\left(\text{Ann }\mathscr{U}\cap\mathscr{V}\right)$ then $\varphi\left(\{x_1,\cdots,x_m,y_1,\cdots,y_k\}\right)=\{0\}$, but since the zero functional is the unique functional with this property we may conclude that $\varphi=\bold{0}$. We, once again, appeal to a dimension argument to finish. Namely that since $\text{Ann }\mathscr{U}\cap\text{Ann }\mathscr{V}=\{\bold{0}\}$ we see that

$\dim_F\text{Ann }\mathscr{U}+\dim_F\text{Ann }\mathscr{V}=\dim_F\mathscr{W}$

and so the conclusion follows. $\blacksquare$

Remark: While the same comment regarding the dimension argument may apply, it does so with less oomph. If $\{\varphi_1,\cdots,\varphi_m,\varphi_{m+1},\cdots,\varphi_k\}$ then a quick check gives that the first $m$ are a basis for $\text{Ann }\mathscr{V}$ and the last $k$ a basis for $\text{Ann }\mathscr{U}$, 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 $\mathscr{W}$.

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.