Problem from Halmos's Finite Dimensional Vector Spaces The following problem was taken from Halmos's Finite Dimensional Vector Spaces:
Let $(a_0, a_1, a_2, \ldots)$ be an arbitrary sequence of complex numbers. Let $x \in P$, where $P$ is the vector space of all polynomials over $\mathbb{C}$. Write $x(t) = \sum_{i=0}^n \xi_it^i$ and $y(x) = \sum_{i=0}^n \xi_i a_i$. Prove that $y(x)$ is an element of the dual space $P'$ consisting of all linear functionals on $P$ and that every element of $P'$ can be obtained in this manner by a suitable choice of the $a_i$.
Now the first part of the problem is not hard to show as $y$ is the functional that takes some polynomial in the vector space, evaluates it at $1$ and to each $\xi_i$ multiplies an $a_i$, then sums all these numbers together. Then it only remains to check that for such a $y$, 
$y(Ax_1 + bx_2) = Ay(x_1) + By(x_2)$, where $x_1,x_2 \in P$ and $A,B \in \mathbb{C}$.
Now for the second part, I don't know how to prove that every linear functional in the dual space must have the form of $y$ above. I can think of specific examples e.g. integrals of polynomials and see why this is true, but it is plain that this will not suffice.                                                                            
There is a more promising approach I can think of. This may not be correct but if we view the $\xi_i's$ as some "basis" (i.e. prove they are linearly independent) and somehow the linear combination represented by $y(x)$ "spans" the dual space then this may suffice. 
Perhaps even related is the question If $y$ is a non-zero functional on a vector space $V$ and if $\alpha$ is an arbitrary scalar, does there necessarily exist a vector $x \in V$ such that $y(x) = \alpha$?
Please do not leave complete answers as I would like to complete this myself.
$\textbf{Note}$: The notation $y(x)$ does not mean a function evaluated at a point e.g. $y=f(x)=x^2$ but rather a functional $y$ evaluated at a vector $x$ belonging to some vector space.
 A: The following steps lead to a solution:
(1) Prove that the tuple $(1,x,x^2,\dots)$ is a basis for the complex vector space $V$ of all polynomials (in the variable $x$) with coefficients in $\mathbb{C}$.
(2) If $p(x)=\sum_{i=0}^n a_ix^i$ and if $\Lambda$ is a linear functional on $V$, prove that $\Lambda(p(x))=\sum_{i=0}^n a_i\Lambda(x^i)$. (Hint: the linearity of the functional $\Lambda$ is, of course, relevant.)
(3) Therefore, the linear functional $\Lambda$ is completely determined by the values $\Lambda(x^i)$ for $i\geq 0$ an integer.
The following exercises are relevant:
Exercise 1: Prove the Riesz representation theorem for finite dimensional Hilbert spaces:

If $H$ is a finite dimensional Hilbert space and if $\Lambda$ is a linear functional on $H$, then there exists a unique vector $y\in H$ such that $\Lambda(x)=\langle x,y \rangle$ for all $x\in H$. (Note that $\langle \cdot,\cdot \rangle$ is the inner product on $H$.)

(Hint: note that $H$ has an orthonormal basis.)
Exercise 2: Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and let $(v_1,\dots,v_n)$ be a basis of $V$. Prove that if $(a_1,\dots,a_n)$ is a tuple of complex numbers, then the map $\Lambda:V\to\mathbb{C}$ given by $\Lambda(\sum_{i=1} c_iv_i)=\sum_{i=1}^n c_ia_i$ (where $c_i\in\mathbb{C}$ for all $1\leq i\leq n$) is a linear functional on $V$. Conversely, prove that all linear functionals on $V$ have this form. 
Exercise 3: Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and let $(v_1,\dots,v_n)$ be a basis of $V$. If $1\leq i\leq n$, let $\Lambda_i:V\to \mathbb{C}$ be defined by the rule $\Lambda_i(v_i)=1$ and $\Lambda_i(v_j)=0$ if $j\neq i$. Prove that $\Lambda_i$ is a linear functional on $V$ for all $1\leq i\leq n$. If $V^{*}$ is the dual space of $V$ (= vector space of all linear functionals on $V$), prove that $(\Lambda_1,\dots,\Lambda_n)$ is a basis of $V^{*}$.
Exercise 4: Let $V$ be the vector space of all polynomials (in the variable $x$) with coefficients in $\mathbb{C}$. If $i\geq 0$, let $\Lambda_i:V\to \mathbb{C}$ be defined by the rule $\Lambda_i(x^i)=1$ and $\Lambda_i(x^j)=0$ if $j\neq i$. Prove that $\Lambda_i$ is a linear functional on $V$ for all $i\geq 0$. Is the tuple $(\Lambda_0,\Lambda_1,\Lambda_2,\dots)$ a basis of the dual space of $V$? 
I hope this helps!
A: It is not hard to show that for each $i=1,2, \ldots n$, $ \exists! \Lambda_i$ s.t. $\Lambda_i[v_i] = \delta_{ij}$. Now assume that these $\Lambda_i's$ are linearly dependent, viz. there exists scalars $a_1,a_2, \ldots a_n$, not all zero s.t.
$a_1\Lambda_1 + a_2\Lambda_2 \ldots a_n\Lambda_n = 0.$
Let the L.H.S.  operate on $v_1$, so that $a_1 = 0$. Continuing this process for all the vectors in the basis shows that every $a_i$ is zero. It follows that the set of linear functionals $\{\Lambda_1, \Lambda_2 \ldots \Lambda_n\}$ is linearly independent. Now for any $v \in V,$ write $v = \sum_{i=1}^n \xi_i v_i$, where the $\xi_i \in \mathbb{F}$. Then by definition of the $\Lambda_i$, we have
$\Lambda_1[v] = \xi_1, \quad \Lambda_2[v] = \xi_2, \quad \ldots \quad \Lambda_n[v] = \xi_n.$
Next we note that for any $y$ in $V^{*}$
$\begin{eqnarray*} y[v] &=& \xi_1y[v_1] + \xi_2y[v_2] + \ldots \xi_ny[v_n] \\

&=&\Lambda_1[v]y[v_1] + \Lambda_2[v]y[v_2] + \ldots \Lambda_n[v]y[v_n] 

\end{eqnarray*}$
so that any linear functional is a linear combination of the $\Lambda_i's$. Since these $\Lambda_i's$ are linearly independent, it is plain that $\{\Lambda_1, \Lambda_2 \ldots \Lambda_n\}$ is a basis of the dual space $V^{*}$.
