# How to show annihilator has dimension m-n (with Proof)

I would like to show the following:

Given a vector spaces $V$, a subspace $S \subset V$ and an the dual space $V^*$ to $V$. Show that: $$\dim(N)+\dim(S) = \dim(V) = \dim(V^*)$$, where $N \subset V^*$ is the annihilator to $S$. Assume only finite dimensional vector spaces.

I tried a proof but I am not sure if this is correct:

## Proof:

I give a try :-)
Assume that
$\dim(S) = m$,
$\dim(V) = \dim(V^*) = n$
The annihilator is given as: $$N:= \{ \boldsymbol{\alpha} \mid \langle \boldsymbol{\alpha},\mathbf{x} \rangle =0 \quad ,\quad \forall \mathbf{x} \in S \}$$ Where $\langle \boldsymbol{\alpha},x \rangle$ is the duality pairing, it is a bilinear form as: \begin{aligned} \textrm{B}( \boldsymbol{\alpha},\mathbf{x} ) : V^* \times V &\rightarrow \mathbb{R} \\ \boldsymbol{\alpha},\mathbf{x} &\mapsto \langle \boldsymbol{\alpha} ,\mathbf{x} \rangle \end{aligned} We fix basis vectors for the following spaces:
Basis for $V$: $\{\mathbf{e}_i\}$ with $i \in \{1,n\}$
Basis for $S$: $\{\mathbf{e}_i\}$ with $i \in \{1,m\}$
Basis for $V^*$: $\{\boldsymbol{\alpha}^i\}$ with $i \in \{1,n\}$
Now with $\boldsymbol{\beta} =\beta_j \boldsymbol{\alpha}^j \in V^*$ and $\mathbf{x} = x^i \mathbf{e}_i \in V$, it follows that $$\langle \boldsymbol{\beta} ,\mathbf{x} \rangle = \langle \beta_j \boldsymbol{\alpha}^j ,x^i \mathbf{e}_i\rangle = \beta_j \langle \boldsymbol{\alpha}^j , \mathbf{e}_i \rangle x^i = \beta_j \ B^j_{\ i} \ x^i \quad i,j \in \{1,n\}$$

We want now for every $\mathbf{x} \in S$ and a corresponding $\boldsymbol{\alpha} \in N$ that:
\begin{aligned} \langle \boldsymbol{\beta} ,\mathbf{x} \rangle = 0 &= (\beta_1, \cdots, \beta_n)\left( \begin{array}{ccc} B^1_{\ 1} & \cdots & B^1_{\ m} \\ \vdots & \cdots & \vdots \\ \vdots & \cdots & \vdots \\ B^n_{\ 1} & \cdots & B^n_{\ m} \end{array} \right) \left(\begin{array}{c} x^1 \\ \vdots \\ x^m\end{array} \right) \end{aligned} = [\boldsymbol{\beta}]^\top\mathbf{B}[\mathbf{x}] =[\mathbf{x}]^\top\mathbf{B}^\top [\boldsymbol{\beta}], \quad \forall [\mathbf{x}] From where follows: $$\mathbf{B}^\top [\boldsymbol{\beta}] = \mathbf{0}$$ If we assume that $\boldsymbol{\alpha}^i$ is the dual basis vector to $\mathbf{e}_i$ with the property: $$\langle \boldsymbol{\alpha}^i, \mathbf{e}_j \rangle = \delta^i_{\ j}$$, where $\delta^i_{\ j}$ is the Kroenecker Delta. It follows that $\delta^j_{\ i} = B^j_{\ i}$. Thus $[\boldsymbol{\beta}]$ lies in the Nullspace of $N$, $[\boldsymbol{\beta}] \in \text{Null}(\mathbf{B}^\top)$.
The dimension of $\text{dim}(\text{Null}(\mathbf{B}^\top)) = n-m$

which means what now? and how can I construct a basis from the given ones for $N$?

Thanks for the inputs!

• You seem to assume $\dim V < \infty$ in your statement, as otherwise $\dim V = \dim V^*$ won't hold. – martini May 28 '13 at 15:44
• I assume you mean $\dim V<\infty$. The restriction $\phi_{|S}$ of a linear functional $\phi\in V^*$ yields a functional in $S^*$. This gives a surjective linear map $V^*\longrightarrow S^*$ whose nullspace is the annihilator of $S$. Apply the rank-nullity theorem (or the first isomorphism theorem if you prefer). – Julien May 28 '13 at 15:46
• Take a look at this post. – H. R. Aug 9 '16 at 19:47

Let $V$ a finite dimensional vector space with $dim\;V=n$ and $V'$ be its dual space over a field $\Bbb F$.

This can be proved in two ways:

$\mathbf {Proof \;1}:$ Let $B_S=\{s_1,...,s_k\}$ be a basis for the subspace $S$. So $dim\;S=k$. Extend $B_S$ to a basis of $V$ say $B_V=\{s_1,...,s_k,s_{k+1},...,s_n\}$. Let $B^{'}_V=\{f_1,...,f_n\}$ be the dual basis to $B_V$ where $f_i(s_j)=\delta_{i,j}$, the Kronecker Delta.

My Claim is that $\{f_{k+1},...,f_n\}$ is the basis of $N=S^0$. Indeed let $f\in V'$ then $f=a_1f_1+...+a_nf_n$ for some $a_1,...,a_n \in \Bbb F$. Now $f(s_j)=a_jf_j(s_j)=a_j$. So $f=f(s_1)f_1+...+f(s_n)f_n$. If $i\in \{k+1,...,n\}$ and $j \in \{1,...,k\}$ then $f_i(s_j)=0$. So if $s\in S$ then $s=b_1s_1+...+b_ns_k$ for some $b_1,....,b_k\in \Bbb F$. Then $f_i(s)=0\; \forall i \in \{k+1,...,n\}$. Hence $f_{k+1},...,f_n \in S^0$. Then $span \{f_{k+1},...,f_n \}\subseteq S^0$ since $S^0$ is a subspace. Now suppose $f\in S^0$ then $f(s_1)=...=f(s_k)=0$ since $s_1,...,s_k \in S$. Then writing $f=f(s_1)f_1+...+f(s_n)f_n$, the first $k$ terms are zero i.e $f=f(s_{k+1})f_{k+1}+...+f(s_n)f_n$. So $f\in span\{f_{k+1},...,f_n\}$. Hence $S^0\subseteq span\{f_{k+1},...,f_n\}$. Hence $S^0=span\{f_{k+1},...,f_n\}$, and of course $\{f_{k+1},...,f_n\}$ is linearly independent being a subset of a set of basis vectors.So $dim\;S^0=n-k=dim\;V-dim\;S$.So $dim\;V=dim\;S + dim\;S^0$ . $\lhd$

$\mathbf {Proof \;2}:$ Now suppose $i:S\rightarrow V$ be the natural inclusion map i.e $i(s)=s,\; \forall s \in S$. Consider its dual map $i':V'\rightarrow S'$ defined as follows $i'(\phi)=\phi\,\circ i$. Thus $i'$ is a linear map from $V'$ to $S'$. Then $dim\,range(i')+dim\,null(i')=dim\,V'=dim\,V$. Now my first claim is,

$\it{Claim \;1}:$ $null(i')=S^0.$

$\it{Pf}:$ If $\phi \in null(i')\Leftrightarrow i'(\phi)=0\Leftrightarrow \phi\,\circ i=0\Leftrightarrow(\phi\,\circ i)(s)=0,\,\forall s\in S\Leftrightarrow\phi(i(s))=0,\,\forall s\in S$ $\Leftrightarrow\phi(s)=0,\, \forall s\in S\Leftrightarrow \phi \in S^0\,\bullet$

Now we have $dim\,range(i')+dim\,S^0=dim\,V$. This means that $dim\,range(i')$ must be equal to $dim\,S=dim\,S'$. But $range\,(i')\subseteq S'$. So it means that $i'$ must be surjective i.e $range(i')=S'$. Indeed,

$\it{Claim\;2}:$ $range(i')=S'$.

$\it{Pf}:$ Let $\phi \in S'$. So $\phi:S\rightarrow \Bbb F$. We extend $\phi$ to a functional $\psi :V\rightarrow \Bbb F$ such that $\psi(s)=\phi(s),\,\forall s\in S$. So $i'(\psi)=\psi\,\circ i$. But $(\psi\, \circ i)(s)=\psi(i(s))=\psi(s)=\phi(s),\,\forall s \in S$. Hence $\psi\, \circ i = \phi \Leftrightarrow i'(\psi)=\phi$. So $S'=range(i')\,\bullet$.

Finally we have $dim\,S'+dim\,S^0=dim\,V\Rightarrow dim\,S+dim\,S^0=dim\,V.\,\lhd$