How prove this $\sigma_{2}=\sigma_{1}\sigma\Longleftrightarrow \ker(\sigma_{1})\subset\sigma_{2}$

Question

let $\sigma_{1},\sigma_{2}$ be a linear transformation from an n-dimension vector space $V$,show that follow two condition are equivalent

$$(1):\ker(\sigma_{1})\subset\ker(\sigma_{2})$$

(2)there exist transformation $\sigma$ from $V$,such $$\sigma_{2}=\sigma \sigma_{1}$$

My try: since $$\ker(\sigma_{1})\subset\ker(\sigma_{2})\Longrightarrow \dim(ker(\sigma_{1}))<\dim(\ker(\sigma_{2}))$$ then I can't,Thank you

possible duplicate of $\ker T\subset \ker S\Rightarrow S=rT$ when $S$ and $T$ are linear functionals — user37238, Jan 02 '14 at 08:56

score 0 · Answer 1 · answered Jan 02 '14 at 09:39

Suppose that $\sigma_2 = \sigma \sigma_1$. Then for all $x \in \ker \sigma_1$, one has $\sigma_2(x) = \sigma(\sigma_1(x))=\sigma(0)=0$. This proves that $\ker \sigma_1 \subset \ker \sigma_2$.
Suppose that $\ker \sigma_1 \subset \ker \sigma_2$ and let $S$ be a subspace of $V$ such that $S \oplus \ker \sigma_1= V$. The map $f \colon S \to \sigma_1(V)$ defined by $f(x) = \sigma_1(x)$ for all $x \in S$ is an isomorphism (check this). Let $\sigma \in \mathcal{L}(V)$ be an arbitrary endomorphism (check the existence) such that $$\forall x \in S,\quad\sigma(x) = \sigma_2(f^{-1}(x)),$$ so that $\sigma\sigma_1(x)=\sigma_2(x)$ for all $x \in S$. Finally, notice that the assumption $\ker \sigma_1 \subset \ker \sigma_2$ implies that $\sigma\sigma_1(x)=\sigma_2(x)$ for all $x \in \ker \sigma_1$. By linearity, we conclude that $\sigma_2 = \sigma\sigma_1$ because $S \oplus \ker \sigma_1= V$.
Actually, under condition (1), the non-empty set $\{\sigma \in \mathcal{L}(V) : \sigma_2 = \sigma\sigma_1\}$ is an affine subspace of $\mathcal{L}(V)$ of dimension $(\dim \ker \sigma_1)^2$.

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