Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional

Theorem
Let $V$ be a complex vector space.

Let $\map \LL {V, \C}$ be the space of all linear transformations from $V$ to complex numbers $\C$.

Let $\ell, L \in \map \LL {V, \C}$ be such that $\ker \ell \subseteq \ker L$ where $\ker$ denotes the kernel.

Then:


 * $\exists c \in \C : L = c \ell$

Proof
Suppose $\ell = \mathbf 0$.

Then $\ker \ell = V$, i.e. the kernel of $\ell$ is the entire vector space $V$.

Moreover:


 * $\ker \ell \subseteq \ker L \implies \ker L = V$

Therefore $L = \mathbf 0$, and we can set $c = 0$ to have:


 * $L = \mathbf 0 = 0 \cdot \ell$

Suppose $\ell \ne \mathbf 0$.

By Linear Transformation Maps Zero Vector to Zero Vector:


 * $\exists v_0 \in V : v_0 \ne \mathbf 0 : \map \ell {v_0} \ne 0$

Let $v \in V$.

Let:


 * $w = v - c_v v_0$

where $c_v \in \C$.

Then:

Suppose $w \in \ker \ell$.

Then $w \in \ker L$ and:


 * $\map \ell w = \map L w = 0$.

Now we will seek for $w$ such that $\map \ell w = 0$.

Hence, if $\ds c_v = \frac {\map \ell v }{\map \ell {v_0} }$ then:

Therefore:


 * $\ds c = \frac {\map L {v_0} }{\map \ell {v_0} }$