Given $k$ distinct linear operators, prove such an $\alpha$ exists

Question

I have $k$ distinct linear operators $\{\phi_i\}$ which act on $V$, a vector space on some number field $K$ (in the sense that $\Bbb Q$ is the smallest possible one). Now I have to prove that there exists some $\alpha\in V$ such that all the $\{\phi_i(\alpha)\}$ are distinct.

My thoughts are as follows:

For any pair $\phi_i,\phi_j$, I can find at least a one-dimensional subspace (say $\text{span}(u)$) which fulfils $\phi_i(v)\ne\phi_j(v)$ as long as $v\in\text{span}(u)-\{0\}$. I cannot guarantee higher dimensions, because for example, $A=\text{diag}(1,0,\cdots,0)$ and $B=\text{diag}(2,0,\cdots,0)$ differ only in $\text{span}(e_1)$.

So I will get $\frac{k(k-1)}{2}$ such pairs and hence $\frac{k(k-1)}{2}$ such $u$s. I want to construct a fulfilling $\alpha$ directly from these $u$s but I don't know how.

Can you help me? Thanks!

Answer 1

Given $\phi_1$ and $\phi_2$ distinct linear operators on $V$ we have $\phi_1(v)=\phi_2(v)\iff \phi_1(v)-\phi_2(v)=0$. So the set of elements such that $\phi_1(v)=\phi_2(v)$ is $Ker(\phi_1-\phi_2)$, a proper subspace of $V$.

What you want to prove is:

$$\bigcup\limits_{1\leq i<j\leq n}Ker(\phi_i-\phi_j)\neq V$$.

But what we have is a finite union of proper subspaces of a vector space $V$ over an infinite field. Such unions are always proper subsets of $V$, as desired. An excellent proof of this can be found in page $36$ of Roman's Advanced Linear Algebra.