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

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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:

\(\ds \map L v\) \(=\) \(\ds \map L {c_v v_0 + w}\)
\(\ds \) \(=\) \(\ds c_v \map L {v_0} + \map L w\) Definition of Linear Transformation

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 $\map \ell w = 0$.

\(\ds \map \ell w\) \(=\) \(\ds \map \ell v - c_v \map \ell {v_0}\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds c_v\) \(=\) \(\ds \frac {\map \ell v }{\map \ell {v_0} }\)

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

\(\ds \map L v\) \(=\) \(\ds \frac {\map \ell v }{\map \ell {v_0} } \map L {v_0}\)
\(\ds \) \(=\) \(\ds \frac {\map L {v_0} }{\map \ell {v_0} } \map \ell v\)

Therefore:

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


$\blacksquare$


Sources

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