TITLE: Proving that linear functionals are linearly independent if and only if their kernels are different
QUESTION [9 upvotes]: Here is a problem I'm trying to solve:
Let's assume that $\dim X < \infty$. Show that 2 non-zero linear functionals $a^*, b^* \in X^*$ are linearly independent if and only if $\ker a^* \neq \ker b^*$.
I think I managed to do the implication from left side to right side:
($\rightarrow$) $a^*, b^*$ are linearly independent if and only if for all $x \in X$ we have $\alpha a^*(x) + \beta b^*(x) = 0$ if and only if $\alpha=\beta=0$ where $\alpha, \beta$ are scalars. But if $\ker a^* = \ker b^*$, then for some $x' \in \ker a^*$ we have $a^*(x') = b^*(x')=0$ so for $\alpha=\beta=1$ we get $a^*(x') + b^*(x') = 0$ which is a contradition. So $\ker a^* \neq \ker b^*$
But I really don't know where to start when trying to prove the implication in the other way. The information that $\ker a^* \neq \ker b^*$ seems like it's not enough to say whether $a^*,b^*$ are linearly independent. So, is there a way to prove that using only that information?

REPLY [6 votes]: Suppose $\text{ker}( a^{\star}) \neq \text{ker}(b^{\star})$. If $a^{\star}$ and $b^{\star}$ are linearly dependent, there are $c$, $d$ such that $c \cdot a^{\star} (v) + d\cdot b^{\star}(v) = 0$ for all $v\in X$. Choose $w\in \text{ker}( a^{\star}) $. Then $$d\cdot b^{\star}(w) = c\cdot a^{\star} (w) + d\cdot b^{\star}(w) = 0$$ so $w\in \text{ker}( b^{\star}) $. So $\text{ker}( a^{\star}) \subset \text{ker}( b^{\star}) $. Similarly, we have the other inclusion. Thus $\text{ker}( a^{\star}) = \text{ker}( b^{\star})$, contradiction.