Prove fact in abstract algebra about Homomorphism in vector space.

Question

Let $V$ will be vector space under field $K$. And also $K$ is a vector space under $K$. We have two Homomorphism's $l_1,l_2 \in \hom(V,K)$ and also $\ker(l_1) = \ker(l_2)$. How to prove that $\exists \alpha \in K (\alpha\ne 0)$ that $l_1 = \alpha l_2$.

Answer 1

The answer by @D_S is very nice in the finite-dimensional situation. The result can be extended to arbitrary dimensions as follows:

Suppose that $v\in V$ such that $l_1(v)=a_1\not=0$. Then $v\not\in\ker(l_1)$. Since $\ker(l_1)=\ker(l_2)$, $v\not\in\ker(l_2)$. Therefore, $l_2(v)=a_2\not=0$. Let $\alpha=\frac{a_1}{a_2}$.

Let $w\in V$ be such that $l_1(w)=b_1\not=0$. Then, as above $w\not\in\ker(l_1)$ so $w\not\in\ker(l_2)$. Let $l_2(w)=b_2\not=0$.

Observe that $v-\frac{a_1}{b_1}w$ is in the kernel of $l_1$ since $$ l_1(v-\frac{a_1}{b_1}w)=l_1(v)-\frac{a_1}{b_1}l_1(w)=a_1-\frac{a_1}{b_1}b_1=0. $$ Therefore, $v-\frac{a_1}{b_1}w$ is in the kernel of $l_2$. Therefore, $$ 0=l_2(v-\frac{a_1}{b_1}w)=l_2(v)-\frac{a_1}{b_1}l_2(w)=a_2-\frac{a_1}{b_1}b_2. $$ Therefore, $b_2=\frac{a_2}{a_1}b_1$. In other words, $b_1=\alpha b_2$.

Therefore, for all $w\not\in\ker(l_1)$, $l_1(w)=\alpha l_2(w)$. Also, for all $w\in\ker(l_1)$, $w\in\ker(l_2)$, so $l_1(w)=0=\alpha l_2(w)$.

Answer 2

Let $n$ be the dimension of $V$. By the rank nullity theorem, the dimension of the kernels of $l_1$ and $l_2$ is $n-1$. Let $v_1, ... , v_{n-1}$ be a basis for $\textrm{Ker}(l_1) = \textrm{Ker}(l_2)$, and let $v_n$ be an element not in the span of $v_1, ... , v_{n-1}$. Then $v_1, ... , v_n$ is a basis for $V$, and $l_i$ ($i = 1,2$) is completely determined by its effect on $v_n$:

$$l_i(c_1v_1 + \cdots + c_nv_n) = l_i(c_1v_1 + \cdots + c_{n-1}v_{n-1}) + l_i(c_nv_n) = 0 + c_nl_i(v_n)$$

Since $l_1(v_n), l_2(v_n)$ are nonzero, choose $0 \neq \alpha$ so that $l_1(v_n) = \alpha l_2(v_n)$.