Prove linear independence of $1+x^3-x^5, 1-x^3,1+x^5$ 
Prove linear independence of $1+x^3-x^5,1-x^3,1+x^5$ in the Vector Space of Polynomials

The attempts I found online all are quite easy. You just substitute something in for $x$ into the equation $a(1+x^3-x^5)+b(1-x^3)+c(1+x^5)=0$ for example $x=1,0,-1$ and this will give you three equations where you can show that $a,b,c=0$. But why can we substitute something in? If I define the Vector Space of Polynomials in a very abstract way that $\sum_{i} \alpha_i x^{i}+\sum_{i} \beta_{i} x^{i}:=\sum_{i} (\alpha_{i}+\beta_{i})x^{i})$ and $(\sum_{i}^{n} \alpha_i x^{i})(\sum_{i}^{m} \alpha_{i} x^{i} ):=\sum_{i=0}^{n+m} c_i x^i$ with $c_k=a_0 b_k+a_1 b_{k-1}+...+a_{k} b_0$ and a $x$ is just an abstract symbol with absolutely no meaning why should one be allowed to substitute something for $x$ or even worse differentiate the equation?
 A: In fact your independence condition can be rewritten $(c-a)x^5+(a-b)x^3+(a+b+c)=0$
It has to be true for all $x$ so it is equivalent to solve the system $\begin{cases}c-a=0\\a-b=0\\a+b+c=0\end{cases}$
And the solution is obviously $a=b=c=0$.
Also in $\mathbb R$, this polynomial has at most $5$ roots, therefore if you plug $6$ numbers you are assured to force the coefficients, but I feel it is simpler to just identify the polynomial with the null polynomial.
A: In order to deal with this problem, you deal with expressions of the type$$a(1+x^3-x^5)+b(1-x^3)+c(1+x^5).\tag1$$Now, take $\alpha\in\Bbb R$. Then the map$$\begin{array}{rccc}\operatorname{ev}_\alpha\colon&\Bbb R[x]&\longrightarrow&\Bbb R\\&P(x)&\mapsto&P(\alpha)\end{array}$$is a linear map and therefore it maps $(1)$ into $a(1+\alpha^3-\alpha^5)+b(1-\alpha^3)+c(1+\alpha^5)$. But this is a real number now, and so you can do anything that you would do with ordinary numbers.
Besides, if, when $P(x)=a_0+a_x+a_2x^2+\cdots+a_nx^n$, you define $P'(x)=a_1+2a_2x+\cdots+na_nx^{n-1}$, then$$\begin{array}{ccc}\Bbb R[x]&\longrightarrow&\Bbb R[x]\\P(x)&\mapsto&P'(x)\end{array}$$is also a linear map. So, again, you can see what it maps $(1)$ into.
A: The reason that you can substitute and differentiate, is because every polynomial defines a function from $\Bbb{R}$ to $\Bbb{R}$. The set $\operatorname{Map}(\Bbb{R},\Bbb{R})$ of all functions from $\Bbb{R}$ to $\Bbb{R}$ is of course an $\Bbb{R}$-vector space, and the map
$$\Bbb{R}[X]\ \longrightarrow\ \operatorname{Map}(\Bbb{R},\Bbb{R}):\ P(X)\ \longmapsto\ (x\ \longmapsto\ P(x)),$$
is injective. So if the functions defined by these polynomials are linearly independent in $\operatorname{Map}(\Bbb{R},\Bbb{R})$, then also the polynomials are linearly independent in $\Bbb{R}[X]$.
A: Evaluating a polynomial at a point $p$ is a ring homomorphism whose kernel is the ideal generated by $x-p$.
A ring homomorphism always takes zero to zero, so if you have a polynomial equal to zero, and apply the evaluation map, the result will still be zero.
You can choose whatever value you like (for example a value which eliminates one of more of the constants) and get a valid expression equal to zero. It can be a surprisingly powerful technique.
A: Let's play a little with notation: let $$\begin{align}&p_1\colon x\mapsto 1+x^3-x^5,\\&p_2\colon x\mapsto 1-x^3\text{ and}\\&p_3\colon x\mapsto 1+x^5.\end{align}$$
What you want to show is that, for $a,b,c\in\mathbb{R}$
$$ap_1+bp_2+bp_3=\mathbf{0}\implies \begin{bmatrix}a\\ b\\ c\end{bmatrix}=0$$
where $\mathbf{0}$ is the function $\mathbf{0}\colon x\mapsto 0$.
Since $\mathbb{R}[x]$ is a vector space, $P:=ap_1+bp_2+bp_3\in \mathbb{R}[x]$.
Now, what you want to show is that
$$P=\mathbf{0}\implies \begin{bmatrix}a\\ b\\ c\end{bmatrix}=0$$
But $P=\mathbf{0}\iff \forall x\in \mathbb{R},\; P(x)=0$.
Clearly, if $a\in \mathbb{R}$, then  we have that $P(a)=0$ and we can substitute values of $x$.
For differentiation, since $P=\mathbf{0}$, then they must have the same derivative, so $P'(x)=\mathbf{0}$ and we can differentiate.
