$f(\alpha _I) \ne 0$

Question

I need help in this question...

Let $F$ be a field of characteristic zero and let $V$ be a finite dimensional vector space over field $F$. If $\alpha _1,\dots , \alpha_m$ are finitely many vectors in $V$ , prove that there is a linear functional $f$ on $V$ such that $$f(\alpha _i ) \ne 0 \space \forall \space i\in 1,\dots , m .$$

I will be very grateful if someone gives step by step solution to the above problem.. Thanks in advance.

I have two more specific doubts..

$1)$ What is meant by characteristic zero?

$2)$ Can I define $m$ linear functionals $f_1, \dots ,f_m$ in the following way (inspite of the fact that $\alpha _1,\dots , \alpha_m$ need not be a basis.) $$f_i (\alpha _j)=\delta _{i,j}$$ where $\delta_{i,j}$ is the Kronecker delta function ?

Please help me if I am wrong...

Answer 1

1) Characteristic zero means that if you take the multiplicative identity element $\mathbf 1 $ of the field, there is no $n$ such that $\sum_{i=1}^n \mathbf 1 = 0$, i.e. you can't sum it a finite times to end up with the additive identity element. As pointed out in the comments, the fields usually used in linear algebra, namely $\mathbb R, \mathbb C, \mathbb Q$ have characteristic zero. An example of a field with doesn't have characteristic zero is the field consisting of two elements $\{0,1\}$ where $1 + 1 = 0$.

2)Yes, you can do this, as long as you assume additionally that $\alpha_j \neq 0$ for all $j$ and take care of the fact that linear dependence can cause headaches.

Let $\beta_1 \dots \beta_{k}$ be a basis of $\operatorname{span}\{\alpha_1, \dots \alpha_m\}$. An alternative way to define the (same) $f_j$ would be:

$f_j(\alpha) = \begin{cases}\mu & \mbox{if } \alpha = \mu\beta_j \mbox{ for some } \mu \in F\\0 & \mbox{otherwise}.\end{cases}$.

These are indeed linear functionals. Hence, also $$f = \sum_{i=1}^k f_i$$

is a linear functional. We have for $\alpha = \sum_{j=1}^k \mu_j\beta_j:$

$$f(\alpha) = \sum_{j=1}^k \mu_j f_j(\beta_j)=\sum_{j=1}^k \mu_j.$$

Why is this sum $\neq 0$ for $\alpha = \alpha_j$? Is the characteristic $0$ important here?

What's $f(0)$ for any functional?

What would happen if we hadn't done the step with beta? How would $f$ look like in the case for $\alpha_1 = -\alpha_2$ (and $n=2$)?