Dtyjlui

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
$$ langle boldsymbol{beta} ,mathbf{x} rangle = langle beta_j boldsymbol{alpha}^j ,x^i mathbf{e}_irangle = beta_j langle boldsymbol{alpha}^j , mathbf{e}_i rangle x^i = beta_j B^j_{ i} x^i quad i,j in {1,n}$$

We want now for every $mathbf{x} in S$ and a corresponding $boldsymbol{alpha} in N$ that:

$$begin{aligned}
langle boldsymbol{beta} ,mathbf{x} rangle = 0 &= (beta_1, cdots, beta_n)left( begin{array}{ccc}
B^1_{ 1} & cdots & B^1_{ m} \
vdots & cdots & vdots \
vdots & cdots & vdots \
B^n_{ 1} & cdots & B^n_{ m}
end{array} right) left(begin{array}{c} x^1 \ vdots \ x^mend{array} right)
end{aligned} = [boldsymbol{beta}]^topmathbf{B}[mathbf{x}] =[mathbf{x}]^topmathbf{B}^top [boldsymbol{beta}], quad forall [mathbf{x}]
$$
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!

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 $fin V'$ 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 $iin {k+1,...,n}$ and $j in {1,...,k}$ then $f_i(s_j)=0$. So if $sin S$ then $s=b_1s_1+...+b_ns_k$ for some $b_1,....,b_kin 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 $fin S^0$ 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 $fin span{f_{k+1},...,f_n}$. Hence $S^0subseteq span{f_{k+1},...,f_n}$. 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 $i:Srightarrow V$ 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 $phi in null(i')Leftrightarrow i'(phi)=0Leftrightarrow phi,circ i=0Leftrightarrow(phi,circ i)(s)=0,,forall sin SLeftrightarrowphi(i(s))=0,,forall sin S$ $Leftrightarrowphi(s)=0,, forall sin SLeftrightarrow phi in S^0,bullet$

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 $phi:Srightarrow Bbb F$. We extend $phi $ to a functional $psi :Vrightarrow Bbb F$ such that $psi(s)=phi(s),,forall sin S$. 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 $dim,S'+dim,S^0=dim,VRightarrow dim,S+dim,S^0=dim,V.,lhd$