Algebra (Matrix Theory) Linear Maps I had a problem in my booked i tried to prove. Here is the problem
"Let $x_1,\ldots,x_n$ be different real numbers and $y_1,\ldots,y_n,s_1,\ldots,s_n$ some real numbers. Prove that there exists a polynomial $p(x)$ of a degree less than $2n$ such that $p(x_i)=y_i$ and $p'(x_i)=s_i$ for every $i=1,2,\ldots,n.$"
Here is my attempt:
Let $V= \{ p(x)\in \mathbb{R}[x] \mid \deg(p(x))<2n\}$ $\Rightarrow$
$\exists$ $h(x)\in V$ s.t $\deg(h(x))<n<2n$ from here we now can apply Lagrange Interpolation theorem. $\Rightarrow$ $h(x_i)=y_i$.
Let $Q= \{ g(x)\in \mathbb{R}[x] \mid g(x)=p'(x),\ p(x) \in V, \deg(p(x)) < n \}$
where $\dim(Q)=n-1$.
Now let $g(x)\in Q \Rightarrow \deg(g(x))<n-1$
Let $A: Q\rightarrow \mathbb{R}^n$
$g(x) \rightarrow (g(x_1),\ldots,g(x_n))$
Now assume $g(x)\in \ker(A) \Rightarrow g(x_i) = 0$, $n$ zeros $\Rightarrow \ g(x)=0 \ \Rightarrow \ker(A) = \{0\} \Rightarrow A$ injective $+$surjective $\Rightarrow g(x_i)=s_i$
But since $\deg(h(x))<n \Rightarrow h'(x) \in Q \Rightarrow g(x)=h'(x) \Rightarrow g(x_i)=h'(x_i)=s_i$
My Professor had i quick look and stated that this was a good attempt but it was not correct. He said that my proof does not with certainty show that this $h'(x_i)=s_i$ ill get in the end will fullfill $h(x_i)=y_i$
I didnt understand him, because in my opinion im sure that the derivitive of $h(x)$ which fullfill $h(x_i)=y_i$ lies in $Q$ and therefor i can let my $g(x)$ be equal to $h'(x)$
I would be grateful if someone can explain this?
 A: At first I didn't understand your proof at all, but here goes.
1) In the definition of $Q$, it is the set of all $G(x)$, but then you don't mention $G$ again, although I suppose we can assume it is a typo for $g(x)$. So essentially as far as I can tell, $Q = \{p'(x) : \deg p < n-1\}$. Since $V$ contains all polynomials of degree less than $2n$, if we say $\deg p < n-1$, then of course $p \in V$. But now notice that $p\mapsto p'$ has a one dimensional kernel, so that $Q$ has dimension $n-2$ not dimension $n-1$.
2) Then you define the map $A:Q \to \mathbb{R}^n$. You correctly show that the only polynomial in the kernel is 0, although you then say that $\ker A = \varnothing$, which is impossible, the kernel always contains the identity (0). Then you say that this implies $A$ is injective and surjective, but although $A$ is injective, it cannot be surjective, since $\dim Q=n-2< \dim\mathbb{R}^n=n$. Even if $\dim Q$ were $n-1$, this would still be impossible.
3) You then say that since $\deg h < n$, $h' \in Q$. Which makes me believe you meant $Q =\{ p' : \deg p < n\}$, which would be a vector space of dimension $n-1$, which does not invalidate the problem in (2). But you then claim that since $h'\in Q$, the $g(x)$ you found before must be $h'(x)$. This step makes no sense. Why on earth would these two different things be equal? Ah I see, your final question explains your thought process. The answer is that $g$ is a specific polynomial, not a variable. You have found a specific polynomial in $Q$ (or would have if that part had been correct) for which $g(x_i)=s_i$. Unfortunately since this is a specific polynomial, you cannot just let it equal something else.
Finally, if you would like a hint about how to approach the problem, think about Lagrange interpolation, and in particular think about the following polynomials:
$$ E_i = \frac{\prod_{j=1}^n (x-x_j)^2}{x-x_i} = (x-x_1)^2(x-x_2)^2\cdots (x-x_{i-1})^2(x-x_i)(x-x_{i+1})^2\cdots (x-x_n)^2$$
What are their derivatives at the various $x_i$?
A: Define $T:\mathbb{R}[x]\rightarrow \mathbb{R}^{2n}$, $T(p(x))=(p(x_1),\ldots,p(x_n), p'(x_1),\ldots,p'(x_n))$. 
Notice that if $p(x)\in \ker(T)$ then $x_1,\ldots,x_n$ are roots of $p(x)$ with multiplicity bigger or equal to $2$, since they are distinct the degree of $p(x)$ is bigger than 2n-1. 
So if $P_{2n-1}$ is the subspace of polynomials with degree smaller or equal to 2n-1 then $T:P_{2n-1}\rightarrow \mathbb{R}^{2n}$ is injective and therefore surjective.
