# Kernel and rank involving composition of linear transformations

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!

• 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 $S_1\subseteq S_2$. Oct 29, 2019 at 16:47
• @Ted Shifrin I appreciate the comment, but I am truly lost. I can't think why it must be a subset. Any more pointers? Oct 29, 2019 at 16:50
• 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 $x\in\ker(T\circ S)$? Oct 29, 2019 at 16:51
• 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? Oct 29, 2019 at 16:53

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

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

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 $$rk(T∘S)≤min(r(T),r(S))$$