Let $V$ be a vector space, $V^*$ the dual vector space; that is, the vector space of all functionals on $V^*$.
Given $S\subseteq V$, we define the annihilator of $S$ to be $$S^0 = \{f\in V^*\mid f(s)=0\text{ for all }s\in S\}.$$
Given $T\subseteq V^*$, we define $N(T)$ to be the set
$$N(T)=\{v\in V\mid f(v)=0\text{ for all }f\in T\}.$$
You are asking to prove that $\mathrm{span}(T)=(N(T))^0$. Note that trivially $\mathrm{span}(T)\subseteq (N(T))^0$.
The equality does not necessarily hold when $V$ is not finite dimensional. For example, if we take the vector space $V$ of almost null sequences in $\mathbb{R}$, and let $f_i$ be defined by $f_i((a_n))=a_i$, and $T=\{f_n\mid n\geq 0\}$, then $N(T)=\{0\}$; but whereas $V^*$ is uncountably dimensional, so $(N(T))^0=V^*$ is uncountably dimensional, $\mathrm{span}(T)$ is countably dimensional and so $\mathrm{span}(T)\neq (N(T))^0$.
But in the finite dimensional case we are fine. I originally wrote the following solution, which invokes dual bases and the fact that a finite dimensional vector space is canonically isomorphic to its double dual. A solution without invoking these facts follows at the end.
Let $T=\{f_1,\ldots,f_n\}$. We may assume that $f_1,\ldots,f_n$ are linearly independent; let $f_{n+1},\ldots,f_m$ complete $T$ to a basis for $V^*$. Let $g_1,\ldots,g_m$ be the dual basis to $\{f_1,\ldots,f_m\}$ in $V^{**}$. That is, $g_i(f_j) = \delta_{ij}$. Since $V$ is finite dimensional, there is a canonical isomorphism $V\cong V^{**}$ given by $x\mapsto\overline{x}$, where $\overline{x}(f) = f(x)$, “evaluate at $x$”. Thus, there exist vectors $x_1,\ldots,x_m$ such that $f_i(x_j)=\delta_{ij}$ (namely, the vectors corresponding to $g_1,\ldots,g_m$ under this isomorphism).
I claim that $N(T)=\mathrm{span}(x_{n+1},\ldots,x_m)$. Indeed, note that $f_i(x_j)=0$ if $1\leq i\leq n$ and $j\geq n+1$, so the span of $\{x_{n_1},\ldots,x_m\}$ is contained in $N(T)$. Conversely, if $x=\alpha_1 x_1+\cdots + \alpha_m x_m$ lies in $N(T)$, then $f_i(x) = \alpha_i=0$ for $i=1,\ldots,n$, so $x\in\mathrm{span}(x_{n+1},\ldots,x_m)$, as claimed.
Now let $g\in V^*$ be a functional such that $N(T)\subseteq N(g)$. Then $g(x_{n+1})=\cdots=g(x_m)=0$. Express $g$ as a linear combination of $f_1,\ldots,f_m$ as
$$g = \beta_1 f_1 + \cdots + \beta_m f_m.$$
Then $g(x_i)=\beta_i$, hence $\beta_{n+1}=\cdots=\beta_m=0$, so $g\in\mathrm{span}(T)$, which is what we wanted to prove.
Therefore, $(N(T))^0 = \mathrm{span}(T)$, as claimed.
Note by contrast that the dimension of $V$ is immaterial to prove the equivalent result that $N(S^0) = \mathrm{span}(S)$. Indeed, $S\subseteq N(S^0)$, so $\mathrm{span}(S)\subseteq N(S^0)$. Conversely, let $\beta$ be a basis for $\mathrm{span}(S)$, and let $x\notin\mathrm{span}(S)$. Then $\beta\cup\{x\}$ is linearly independent, and so we can complete it to a basis $\gamma$. Let $f\in V^*$ be defined by letting it be $1$ on $x$, $0$ on all other vectors of $\gamma$, and extended linearly. Then $f\in S^0$, but $f(x)\neq 0$, so $x\notin N(S^0)$. Thus, $\mathrm{span}(S)= N(S^0)$.
I should note that what we have is what is called a "Galois connection" (or an antitone Galois connection) between $V$ and its dual $V^*$. You can read more about Galois connections in the Wikipedia article or in George Bergman's Invitation to General Algebra and Universal Constructions
I'm not sure if/why they define the duality between "annihilator of a set of vectors" (the collection of all functionals that send all the vectors to zero) and "vectors annihilated by a set of functionals" (the collection of all vectors that are sent to zero by all functionals in the set). It's an elementary but straightforward realization, in the following sense.
We are used to thinking of functions as having a bunch of points, and then using the function to get values, one value per point. This view takes the domain as a collection of things, and a function as a singular object that acts on those things.
But if we consider collections of functions, it turns out we can also go the other way: take a point, and consider a bunch of functions. We can use the point to get values, one per functions, by evaluation. I take the point $x_0$, and given any function $f$, I look at $f(x_0)$. This is itself a function, that takes functions as inputs.
So in the situation at hand, we have a way of going from collections of vectors to values (using a function), or from a collection of functionals to values (using a vector). In both cases we get linear functions (since for functionals, $(f+g)(v) = f(v)+g(v)$ and $(\alpha f)(v) = \alpha( f(v))$). In a sense, vectors act as functionals on the vector space of functionals! So we can talk about "annihilator" of functionals just as we talk about "annihilator" of sets of vectors.
This is often buried in exercises in standard linear algebra books (not necessarily aimed at graduate students; books like the one you are using, or Friedberg/Insel/Spence). It's possible the authors missed the fact they had not explicitly introduced the two notions.
So, the question is whether given a finite dimensional vector space $V$, and a finite (without loss of generality) linearly independent subset $T$ of $V^*$, the annihilator of $N(T)$ is equal to $\mathrm{span}(T)$ in $V^*$, without invoking dual bases or the isomorphism between $V$ and its double dual.
Let $V$ be a finite dimensional vector space, of dimension $n$. If $f$ is a nonzero functional on $V$, then we know (from the Dimension Theorem, aka the Rank/Nullity Theorem) the its nullspace has dimension $n-1$.
Let’s look at the case of a single functional, to see how this will work.
Lemma. Let $f$ and $g$ be nonzero functionals on $V$. Then $\{f,g\}$ is linearly dependent if and only if $N(f)=N(g)$.
Proof. If $\{f,g\}$ is linearly dependent, then one of $f$ and $g$ is a scalar multiple of the other; and since neither is the zero functional, there exists a nonzero scalar $\alpha$ such that $g=\alpha f$. Then $v\in N(g)$ if and only if $v\in N(f)$.
Conversely, suppose $N(f)=N(g)$. Let $\{v_1,\ldots,v_{n-1}\}$ be a basis for $N(f)=N(g)$, and let $v_n\in V$ be a vector not in $N(f)$. Then $f(v_n)\neq 0$ and $g(v_n)\neq 0$, and moreover, $\{v_1,\ldots,v_{n-1},v_n\}$ is a basis for $V$, so $f$ and $g$ are completely determined by what they do on $v_n$ (since we already know what they do on $\{v_1,\ldots,v_{n-1}\}$. Let $f(v_n)=\alpha$, $g(v_n)=\beta$. Then $g(v_n) = \frac{\beta}{\alpha}f(v_n)$. Thus, $g$ and $\frac{\beta}{\alpha}f$ take the same values on the basis $\{v_1,\ldots,v_n\}$, and so are equal. Since $g$ is a scalar multiple of $f$, it follows that $\{f,g\}$ is linearly dependent. $\Box$
This tells us the following:
Corollary. Let $f$ be a functional on $V$. Then $(N(f))^0 = \mathrm{span}(f)$.
Proof. If $f=0$ then $N(f)=V$, and $(N(f))^0$ is the just the zero functional, i.e., $f$, i.e., $\mathrm{span}(f)$. Assume $f\neq 0$.
Let $g\in (N(f))^0$. Then $N(f)\subseteq N(g)$; hence either $N(g)=V$, in which case $g$ is the zero functional and lies in $\mathrm{span}(f)$, or else $N(f)=N(g)$ (because $\dim(N(f))=\dim(V)-1$), in which case $\{f,g\}$ is linearly dependent, both nonzero, hence $g\in\mathrm{span}(f)$. $\Box$
But the key for the general case is going to be going from one to two functionals (equivalently, from characterizing when one functional is in the span of another to characterizing when one functional is in the span of two others).
Suppose $f$ and $g$ are linearly independent functionals on $V$. Then $N(f)$ and $N(g)$ are both of dimension $n-1$, but different. This means that $N(f)\cap N(g)$ has dimension $n-2$: indeed, we have $N(f)+N(g)=V$, so
$$n = \dim(N(f)+N(g))=\dim(N(f))+\dim(N(g)) - \dim(N(f)\cap N(g))$$
so $n=2n-2 -\dim(N(f)\cap N(g))$ yields that the dimension is exactly $n-2$.
Let $\{v_1,\ldots,v_{n-2}\}$ be a basis for $N(f)\cap N(g)$. Since $N(f)\neq N(g)$, there is a vector $v_{n-1}$ in $N(f)$ that is not in $N(g)$. Note that this means that $v_{n-1}$ is not in $N(f)\cap N(g)$. Normalizing, we may assume that $g(v_{n-1}) = 1$. So $f(v_{n-1})=0$, $g(v_{n-1}) = 1$. Symmetrically, we can find a vector $v_n$ such that $f(v_n)=1$ and $g(v_{n}) = 0$. I claim that $\{v_1,\ldots,v_n\}$ is a basis for $V$. To that end, it is enough to show it is linearly independent. We know $\{v_1,\ldots,v_{n-2}\}$ is linearly independent, and that $v_{n-1}\notin\mathrm{span}(v_1,\ldots,v_{n-2})$, so $\{v_1,\ldots,v_{n-1}\}$ is linearly independent (in fact, it is a basis for $N(f)$). Since $v_n\notin N(f)$, then $v_n$ is not in the span of $v_1,\ldots,v_{n-1}$, so we have a basis of $V$ such that:
- $f(v_i)=g(v_i)=0$ for $i=1,\ldots,n-2$.
- $f(v_{n-1})=0$, $g(v_{n-1}) = 1$;
- $f(v_n) = 1$, $g(v_n)=0$.
Now, suppose $h$ is a nonzero functional. If $h=\alpha f+\beta g$ for some scalars $\alpha$ and $\beta$, then $N(f)\cap N(g)\subseteq N(h)$.
Conversely, assume that $N(f)\cap N(g)\subseteq N(h)$. Then $v_1,\ldots,v_{n-2}$ are linearly independent vectors in $N(h)$, Let $\beta=h(v_{n-1})$ and $\alpha=h(v_n)$. Consider $\alpha f+\beta g$. The value at $v_1,\ldots,v_{n-2}$ is $0$; the value at $v_{n-1}$ is $\beta$, and the value at $v_n$ is $\alpha$; that is, $h=\alpha f +\beta g$, so $h$ lies in $\mathrm{span}(f,g)$. Thus, if $f$ and $g$ are linearly independent functionals on $V$, then a functional $h$ lies in $\mathrm{span}(f,g)$ if and only if $N(f)\cap N(g)\subseteq N(h)$.
And so this is the idea: if we have $k$ linearly independent functionals $f_1,\ldots,f_k$, then the dimension of $N(f_1)\cap\cdots\cap N(f_k)$ is $n-k$, and $h$ lies in $\mathrm{span}(f_1,\ldots,f_k)$ if and only if $N(f_1)\cap \cdots \cap N(f_k)\subseteq N(h)$.
Theorem. Let $V$ be a finite dimensional vector space, $\dim(V)=n$, and let $f_1,\ldots,f_k$, $1\leq k\leq n$ be linearly independent functionals on $V$. Then $\dim(N(f_1)\cap\cdots\cap N(f_k)) = n-k$ and a function $h$ lies in $\mathrm{span}(f_1,\ldots,f_k)$ if and only if $N(f_1)\cap\cdots\cap N(f_k) \subseteq N(h)$.
Proof. We proceed by induction on $k$. We have done $k=1,2$. Now suppose $f_1,\ldots,f_{k},f_{k+1}$ are linearly independent functionals. We know that $\dim(N(f_1)\cap\cdots\cap N(f_k)) = n-k$, and that it is not contained in $N(f_{k+1})$. So $\dim(N(f_1)\cap\cdots\cap N(f_k)\cap N(f_{k+1})) \lt n-k$, a nd more over $N(f_1)\cap\cdots\cap N(f_k) + N(f_{k+1}) = n$, since $N(f_{k+1})$ has dimension $n-1$. Thus, in fact, we have
$$\begin{align*}
n &= \dim(N(f_1)\cap\cdots\cap N(f_k) + N(f_{k+1})) \\
&= \dim(N(f_1)\cap\cdots\cap N(f_k)) + \dim(N(f_{k+1}) - \dim(N(f_1)\cap\cdots\cap N(f_k)\cap N(f_{k+1})),
\end{align*}$$
so $n= (n-k)+(n-1) - \dim(N(f_1)\cap\cdots\cap N(f_{k+1})$. Therefore,
$$\dim(N(f_1)\cap\cdots\cap N(f_{k+1})) = n-k-1 = n-(k+1),$$
as required. This establishes the dimension part of the statement.
Now, we just need to show that $h$ lies in $\mathrm{span}(f_1,\ldots,f_{k+1})$ if and only if $N(h)$ contains $N(f_1)\cap\cdots\cap N(f_{k+1})$. The "only if" is clear, so we just need to show the "if". Let $h$ be a functional that contains $N(f_1)\cap\cdots\cap N(f_{k+1})$; we want to show that $h$ is a linear combination of $f_{1},\ldots,f_{k}$.
Let $v_1,\ldots,v_{n-(k+1)}$ be a basis for $N(f_1)\cap\cdots\cap N_(f_{k+1})$. Now, for each $i$ we know that the intersection of all $N(f_j)$ with $j\neq i$ is not contained in $N(f_{i})$ by the induction hypothesis. Therefore, for each $i$ we can find a vector $w_i$ such that $f_i(w_i)=1$ and $f_j(w_i)=0$ for $j\neq i$. For example, since $f_{k+1}$ is linearly independent from $f_1,\ldots,f_k$, we know by induction that $N(f_{k+1})$ does not contain $N(f_1)\cap\cdots\cap N(f_k)$. So pick a vector $w_{k+1}$ in $N(f_1)\cap\cdots\cap N(f_k)$ that is not in $N(f_{k+1})$, and then normalize so that $f(w_{k+1}) = 1$ (that is, multiply by a suitable nonzero scalar). The argument is the same for each index.
So now we have vectors $v_1,\ldots,v_{n-k-1}$, a basis for $N(f_1)\cap\cdots\cap N(f_{k+1})$. We also have vectors $w_1,\ldots,w_{k+1}$. I claim that $v_1,\ldots,v_{n-k-1},w_1,\ldots,w_{k+1}$ is linearly independent, and hence a basis for $V$. Indeed, note that $f_{i}$ takes the value $0$ at $v_1,\ldots,v_{n-k-1}$ and also at $w_1,\ldots,w_{i-1}$, but not at $w_i$; hence $w_i$ cannot be a linear combination of $v_1,\ldots,v_{n-k-1},w_1,\ldots,w_{i-1}$. This holds for each $i$.
So then we have a basis for $V$, and so $h$ is completely determined by its value on $v_1,\ldots,v_{n-k-1}$ (which is $0$, since we are assuming that $N(f_1)\cap\cdots\cap N(f_{k+1})\subseteq N(h)$) and at $w_1,\ldots,w_{k+1}$. Let $\alpha_i = h(w_i)$. I claim that $h=\alpha_1f_1+\cdots+\alpha_{k+1}f_{k+1}$.
Indeed, both functions take the value $0$ at $v_1,\ldots,v_{n-k-1}$; and at $w_j$ we have
$$(\alpha_1f_1+\cdots+\alpha_{k+1}f_{k+1})(w_j) = \sum_{i=1}^{k+1}\alpha_i\delta_{ij} = \alpha_j=h(w_j).$$
Hence, $h$ is a linear combination of $f_1,\ldots,f_{k+1}$, as desired.
This completes the induction, and the Theorem. $\Box$
This proves our desired result: suppose $T$ is a subset of $V^*$ (which is finite dimensional) and let $f_1,\ldots,f_k$ be a basis for $\mathrm{span}(T)$. Then $N(T)=N(f_1)\cap\cdots\cap N(f_k)$. Then $h\in(N(T))^0$ if and only if $N(T)\subseteq N(h)$, if and only if $N(f_1)\cap\cdots\cap N(f_k)\subseteq N(h)$, if and only if $h$ is a linear combination of $f_1,\ldots,f_k$, if and only if $h\in\mathrm{span}(f_1,\ldots,f_k)=\mathrm{span}(T)$. Therefore:
Theorem. Let $V$ be a finite dimensional vector space, let $S$ be a subset of $V$, and let $T$ be a subset of $V^*$. Then:
- $N(S^0) = \mathrm{span}(S)$.
- $(N(T))^0 = \mathrm{span}(T)$.
I’ll add that the construction above is essentially the construction of the “dual basis” of a basis for $V$ which I invoked in the first solution.
Best Answer
Representing $\alpha_j'$ in the ordered basis $(\alpha_1',\dots,\alpha_n')$ will just give you the standard basis vector $e_j$ because $\alpha_j' = 0\alpha_1' + \dots + 1\alpha_j' + \dots + 0 \alpha_n'$. Multiplying $P$ by $e_j$ gives you the $j$-th column of $P$. So equation (i) says that
$$ [\alpha_j']_{\mathcal B} = \begin{bmatrix} P_{1,j} \\ \vdots \\ P_{n,j} \end{bmatrix}. $$
By definition of $[\alpha_j']_{\mathcal B}$, this means that $$ \alpha_j' = P_{1,j}\alpha_1 + \dots + P_{n,j}\alpha_n = \sum_{i = 1}^n P_{i,j} \alpha_i. $$