Scalar Product , Sequences, Dimension This question was asked in my online course  on advanced linear algebra and I was not able to solve it.
So, I am asking for help here.

Question: (a) Denote by <,> , the canonical scalar product in $\mathbb{R}^n$: $<x,y>= \sum_{i=1}^n x_i y_i$. Suppose that we have two sequences $v_1,...,v_k$ and  $w_1,...,w_k$ $\in \mathbb{R}^n$ vectors satisfying  $<v_i, w_j> = 1 $ if i=j and 0 if $i\neq j$ for all $1 \leq i,j \leq n$. Show that $k\leq n$.

(b) Let $E=\mathbb{R}[x]$. Show that $E^*$ is isomorphic to $\mathbb{R}^\mathbb{N}$.
Attempt: (a)I assumed that k>n  in the hope of obtaining a contradiction but I am not able to proceed
(b) I am really sorry but I don't have anything to show as attempt.
Can you please let me know some hints needed to prove this.
 A: For the first claim, assume there exists a non-trivial linear combination $\sum_ia_iv_i=0$, so assume without loss of generality that $\sum_{i=1}^{j+1}a_iv_i=0$, thus $v_{j+1}=\sum_{i=1}^{j}b_iv_i$ with $b_i=-a_i/a_{j+1}$. Then we have $1=<v_{j+1},w_{j+1}>=\sum_{i=1}^{j}b_i<v_{i},w_{j+1}>=0$, a contradicition, so the $v_i$ are linearly independent. Now, well, say that $v_1\dots,v_k$ can be extended to a basis of $\mathbb R^n$, so $k\le n$.
For the second part we assume that $0\in\mathbb N$, briefly recall the polynomial ring $E=\mathbb R[x]$ and the
dual space $E^*$.
Let's take a look at the map $\iota:\mathbb R^{\mathbb N}\rightarrow E^*$, $v\mapsto\varphi_v$, that maps $v$ to the linear form $\varphi_v:\mathbb R[x]\rightarrow\mathbb R$ given by $\varphi_v(x^k)=v_k$.
Conversely, let $\kappa:E^*\rightarrow\mathbb R^{\mathbb N}$, $\varphi\mapsto(\varphi(x^k))_{k\in\mathbb N}$.
To be thorough, we have to show that $\iota$ is well-defined, it's not really obvious that $\varphi_v$ is unique (and it's also instructive). So, let's look at some $P\in\mathbb R[x]$. Recall that $P=\sum_{k=0}^na_kx^k$ for some $(a_k)_k$ and $n$. But then we have $\varphi_v(P)=\sum_{k=0}^na_k\varphi_v(x^k)=\sum_{k=0}^na_kv_k$, so we know what the image is and thus $\varphi_v$ is unique. So, both $\iota$ and $\kappa$ are well-defined. It's also fairly immediate that $\kappa$ is the inverse of $\iota$, using that $\kappa\circ\iota=\mathrm{id}$ and $\iota\circ\kappa=\mathrm{id}$. So, we're left to show that $\iota$ and $\kappa$ are linear. The easier argument is $\kappa(a\varphi+b\psi)=(a\varphi(x^k)+b\psi(x^k))_k=a(\varphi(x^k))_k+b(\psi(x^k))_k=a\kappa(\varphi)+b\kappa(\psi)$, and the slightly more annoying part is $\iota(av+bw)=\varphi_{av+bw}$, where $\varphi_{av+bw}(x^k)=(av+bw)_k=av_k+bw_k=a\varphi_v(x^k)+b\varphi_w(x^k)$, which then again gives $\varphi_{av+bw}(P)=\sum_ka_k(a\varphi_v(x^k)+b\varphi_w(x^k))=a\varphi_v(P)+b\varphi_w(P)$, and finally $\iota(av+bw)=\varphi_{av+bw}=a\varphi_v+b\varphi_w=a\iota(v)+b\iota(w)$.
