A question about how to show two polynomials are not equal

Question

Let $f\in k[x_i,i=0,\ldots ,n]$ be a a polynomial which contains $x_0$ (i.e. polynomial like $f=x_1$ is not allowed), where $k$ is a field with characteristic $0$. We define a map $$Q_a: k[x_i,i=0,\ldots ,n] \to k[x_i,i=1,\ldots ,n]$$ which sends $f(x_0,\ldots,x_n)$ to $f(a_1x_1+\ldots+a_n x_n, x_1,\ldots,x_n)$, where $a=(a_1,\ldots a_n)$ is a parameter.

My question is

if $a\neq a'$, is it true that $Q_a(f)\neq Q_{a'}(f)$?

I think this should be true, but I did not find a good way to show it.

Answer 1

Counterexample: let $f(x_0,x_1,x_2)=x_0^2+x_1^2+x_2^2$. Let $a=(1,-1)$, and $a'=(-1,1)$.

Then $Q_a(f)=Q_{a'}(f)$.

Answer 2

Start from the Leading coefficient,(If two were equal) the coefficient of the highest power of the polynomial must be equal, then the second highest must be equal ...

As for the proof, you may consider use induction on the degree of polynomials.