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Let $S : U \to V$ and $T : V \to W$ be linear mappings between finite-dimensional vector spaces, and let $T\circ S : U \to W$ be their composition. I need to show two things but I don't even know where to start so explanations would be so helpful!

Firstly:

  • Show that $\ker(S) ⊆ \ker(T \circ S)$, and hence deduce that $n(S) ≤ n(T \circ S)$.

Secondly:

  • Using the first 'show that', show that $r(T \circ S) ≤ \min(r(T), r(S))$, where “$\min$” denotes the minimum.

I presume the latter involves that of the rank-nullity theorem but I don't know how to use it. Thank you again!

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  • $\begingroup$ I misread your question earlier, sorry about that $\endgroup$ Oct 29, 2019 at 16:47
  • $\begingroup$ Here's how to get started. How do you show one set is a subset of another? Suppose $x\in\ker(S)$. Why is $x\in\ker(T\circ S)$? For the "hence deduce" you should know something about dimensions of subspaces $S_1$, $S_2$ when $S_1\subseteq S_2$. $\endgroup$ Oct 29, 2019 at 16:47
  • $\begingroup$ @Ted Shifrin I appreciate the comment, but I am truly lost. I can't think why it must be a subset. Any more pointers? $\endgroup$
    – Nipster
    Oct 29, 2019 at 16:50
  • $\begingroup$ I got you started. You need to use definitions. What does it mean to say $x\in\ker(S)$? What does it mean to say $x\in\ker(T\circ S)$? $\endgroup$ Oct 29, 2019 at 16:51
  • $\begingroup$ Well it means that $x$ gets mapped to zero via $S$, I suppose that because it gets mapped to zero it will stay as zero when it gets mapped with $T$ therefore it must be a subset? $\endgroup$
    – Nipster
    Oct 29, 2019 at 16:53

1 Answer 1

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$A \subseteq B \Leftrightarrow A \cap B = A$

  • $ ker(S) \cap ker(T∘S) \subset ker(S)$

Let $ x \in ker(S) \cap ker(T∘S)$, $x \in ker(S)$

  • $ ker(S) \subset ker(S) \cap ker(T∘S)$

Let $ x \in ker(S) $ , $T(S(x)) =T(0) = 0$, so $ x \in ker(S) \cap ker(T∘S)$

Then $$ker(S) \subseteq ker(T∘S) \Leftrightarrow ker(S) \cap ker(T∘S) = ker(S)$$

and $$dim(ker(S)) ≤ dim(ker (T∘S))$$

According to Rank–nullity theorem : $ r(T∘S) +dim(U) =- dim(ker (T∘S))$

and $ r(S) + dim(U) =-dim(ker (S))$

Then $rk(T∘S) ≤ rk(S)$

and $rk(T∘S)≤ rk(T)$ because $ Im(T∘S) \cap Im(S)$,

Let $ y \in Im(T∘S), \exists x \in V / T(S(x)) = y$

therefore $ y \in Im(T)$ because $ S(x) \in V$.

Finally $$rk(T∘S)≤min(r(T),r(S))$$

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