linear Algebra Kernel and Image Proof

Question

let there be S,T linear operators working in vector space $U$; $$ T,S : U \rightarrow U $$

Prove that:$$ Ker(ST)=U \Longleftrightarrow ImT \subseteq Ker(S) $$

Attempt at a Solution:

Left to right Direction:

$ Ker(ST)=U \Rightarrow \forall u\in U :(ST)(u) = \vec{0}_U $ $$ S(T(u))=\vec{0}_U\ $$

$\Rightarrow$ $\forall u\in U, T(u)\in Ker(S)$ $\Rightarrow$ $Im T \subseteq Ker(S)$

Right to left:

$Im T \subseteq Ker(S)$ $\Rightarrow$ $\forall u\in U, T(u)\in Ker(S)$

$\Rightarrow$ $ S(T(u))=\vec{0}_U\ $ $$ \forall u\in U, (ST)(u) = \vec{0}_U $$ $\Rightarrow$ $ U \subseteq Ker(ST) $ $$ \forall v \in Ker(ST) , v\in U $$ $\Rightarrow$ $ Ker(ST) \subseteq U $ $\Rightarrow$ $ U = Ker(ST) $

$QED$

Is this right?

Yep, just read through it too, looks perfect! – Art Gower Sep 16 '13 at 10:46 — Art Gower, Sep 16 '13 at 10:46
Many thanks to all critics and reviewers! – Bak1139 Sep 16 '13 at 13:32 — Bak1139, Sep 16 '13 at 13:32

score 1 · Accepted Answer · answered Oct 08 '13 at 20:59

1

This answer intends to remove the question from the unanswered queue.

As already noted in the comments, your solution looks perfect.

answered Oct 08 '13 at 20:59

Julian Kuelshammer

9,670

linear Algebra Kernel and Image Proof

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