# Linear dependence of linear functionals

Problem: Let V be a vector space over a field F and let $\alpha$ and $\beta$ be linear functionals on $V$. If $\ker(\beta)\subset\ker(\alpha)$, show $\alpha = k\beta$, for some $k\in F$.

A proposed solution is in the answers below.

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If $\alpha$ is the zero functional, we are done, because we take $k=0$.
Otherwise, consider a basis $\{v_\alpha\}$ of $V$. Let $\{v_p\}$ be the vectors that $\beta$ maps to nonzero scalars, and the $\{v_r\}$ the basis vectors mapped to zero. Then $\alpha$ must also map every vector in $\{v_r\}$ to 0, by hypothesis.
If $\{v_p\}$ contains just one vector we are done, because we can just scale $\beta$ so that $\alpha$ and $\beta$ agree on this basis vector. Otherwise, choose two vectors $v_1$ and $v_2$ in this set. Let $\beta(v_1)=b_1$, $\beta(v_2)=b_2$, $\alpha(v_1)=a_1$, and $\alpha(v_2)=a_2$. We want to show that $b_1/a_1=b_2/a_2$. Assume not. Then consider the vector $b_2v_1-b_1v_2$. We see that $\beta$ maps this to 0, but $\alpha$ does not, a contradiction.
Why does $\alpha(v)=\frac{\beta(v)}{\beta(v_1)}$ for all $v$? – Chris Eagle Jun 14 '12 at 9:32
Yes, I think there's a serious flaw: you haven't given any hint as to why your $k$ works. "Linearity" is not an explanation. – Chris Eagle Jun 14 '12 at 9:36
So we have $k\beta(v_1)=\alpha(v_1)$, and $\beta(v_\alpha)=\alpha(v_\alpha)$ for all other basis vectors $v_\alpha$. So for any any vector $v\in V$, we write it as a combination of basis vectors, and we see easily that $k\beta=\alpha$, right? – Potato Jun 14 '12 at 9:39
Why does $\beta(v_\alpha)=\alpha(v_\alpha)$ for all other $v_\alpha$? – Chris Eagle Jun 14 '12 at 9:41
No, we don't. All we specified when we chose the basis is that $\alpha(v_1)=1$. – Chris Eagle Jun 14 '12 at 9:44