Linear Algebra, Polynomials Let $p \in \mathbb{P}_n(\mathbb{C})$ for some $n$ and suppose there exist distinct real numbers $x_0,x_1,...,x_n$ such that $p(x_j)\in \mathbb{R}$ for all $j=0,1,...,n$. Prove that all coefficients of $p$ are real.
 A: Call the unknown coefficients $t_0,t_1,\ldots,t_n$, and the value at the (known) real numbers $x_j$ as $y_j$.
So we get equations of the kind:
$\sum_k x_j^k t_k = y_j$ for each $j$ where every coefficient is real. Solving the linear system in $t_0,t_1,\ldots ,t_n$ will give a real solution.
A: A different approach: You may know that the polynomial which goes through $m+1$ points,namely $(x_0,p(x_0)),(x_1,p(x_1)),$ . . . $ ,(x_m,p(x_m))$,where $x_0,$ . . . $,x_m$ are all distinct from each other, is unique by one of the exercises which are just for the same chapter. Then you should know how to construct a Lagrange Polynomial from these $m+1$ points and because of the uniqueness of $p$ the conclusion is immediate.
A: I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
This problem is the same problem as Exercise 4.10 in this book.
I used Exercise 4.5 in this book to solve Exercise 4.10.
I used 4.12 on p.123 in this book to solve Exercise 4.5.
So, first, I write the statement of 4.12 without a proof.
Next, I solve Exercise 4.5.
Next, I solve Exercise 4.10.

4.12 A polynomial has at most as many zeros as its degree
Suppose $p\in\mathcal{P}(\mathbb{F})$ is a polynomial with degree $m\geq 0$. Then $p$ has at most $m$ distinct zeros in $\mathbb{F}$.


Exercise 4.5:
Suppose $m$ is a nonnegative integer, $z_1,\dots,z_{m+1}$ are distinct elements of $\mathbb{F}$, and $w_1,\dots,w_{m+1}\in\mathbb{F}.$ Prove that there exists a unique polynomial $p\in\mathcal{P}_m(\mathbb{F})$ such that $$p(z_j)=w_j$$ for $j=1,\dots,m+1.$
[This result can be proved without using linear algebra. However, try to find the clearer, shorter proof that uses some linear algebra.]


My solution:
Let $T$ be a mapping from $\mathcal{P}_m(\mathbb{F})$ to $\mathbb{F}^{m+1}$ such that $\mathcal{P}_m(\mathbb{F})\ni p\mapsto Tp=\begin{pmatrix}p(z_1)\\p(z_2)\\\vdots\\p(z_{m+1})\end{pmatrix}\in\mathbb{F}^{m+1}$.
It is easy to verify that $T$ is a linear mapping.
If $p\neq 0$, then by 4.12 $p\notin\operatorname{null}T$.
So, $\operatorname{null}T=\{0\}.$
So, $T$ is injective.
By Fundamental Theorem of Linear Maps on p.63, $\dim\mathcal{P}_{m}(\mathbb{F})=\dim\operatorname{null}T+\dim\operatorname{range}T.$
Since $\operatorname{null}T=\{0\}$, $m+1=\dim\mathcal{P}_{m}(\mathbb{F})=\dim\operatorname{range}T.$
Since $\operatorname{range}T\subset\mathbb{F}^{m+1}$, $\operatorname{range}T=\mathbb{F}^{m+1}$.
So, $T$ is surjective.
So, $T$ is bijective.
So, there exists a unique polynomial $p\in\mathcal{P}_m(\mathbb{F})$ such that $$p(z_j)=w_j$$ for $j=1,\dots,m+1.$


Exercise 4.10:
Suppose $m$ is a nonnegative integer and $p\in\mathcal{P}_m(\mathbb{C})$ is such that there exist distinct real numbers $x_0,x_1,\dots,x_m$ such that $p(x_j)\in\mathbb{R}$ for $j=0,1,\dots,m$. Prove that all the coefficients of $p$ are real.


My solution:
Let $S$ be a mapping from $\mathcal{P}_m(\mathbb{R})$ to $\mathbb{R}^{m+1}$ such that $\mathcal{P}_m(\mathbb{R})\ni q\mapsto Sq=\begin{pmatrix}q(x_1)\\q(x_2)\\\vdots\\q(x_{m+1})\end{pmatrix}\in\mathbb{R}^{m+1}$.
By Exercise 4.5, $S$ is surjective.
So, there exists $q\in\mathcal{P}_m(\mathbb{R})$ such that $q(x_j)=p(x_j)$ for $j=0,1,\dots,m$.
Let $T$ be a mapping from $\mathcal{P}_m(\mathbb{C})$ to $\mathbb{C}^{m+1}$ such that $\mathcal{P}_m(\mathbb{C})\ni r\mapsto Tr=\begin{pmatrix}r(x_1)\\r(x_2)\\\vdots\\r(x_{m+1})\end{pmatrix}\in\mathbb{C}^{m+1}$.
By Exercise 4.5, $T$ is bijective.
So, there exists a unique $r\in\mathcal{P}_m(\mathbb{C})$ such that $r(x_j)=p(x_j)$ for $j=0,1,\dots,m$.
So, $r=p$.
But $q\in\mathcal{P}_m(\mathbb{R})\subset\mathcal{P}_m(\mathbb{C})$ and $q(x_j)=p(x_j)$ for $j=0,1,\dots,m$.
So, $r=q$.
So, $p=q\in\mathcal{P}_m(\mathbb{R})$.

