Rank of a bijective function and bijectivity of a composition

Question

I have problems with two similar questions:

Let $T: \mathbb{R}^n \to \mathbb{R}^m$ and $S: \mathbb{R}^m \to \mathbb{R}^l$ be linear maps. Show that if $S$ is bijective, then $\text{nullity} (S \circ T) = \text{nullity} T$ and $\text{rank} (S \circ T) = \text{rank} T$.

and

Assume $T: \mathbb{R}^n \to \mathbb{R}^n$ and $S: \mathbb{R}^n \to \mathbb{R}^n$ are bijective linear maps. Show that $S \circ T$ is bijective.

With the first question, I can see how to show that $\text{rank} (S \circ T) = \text{rank} T$ using the fact $\text{Im} S = \mathbb{R}^m$, but I don't understand how to get to $\text{nullity} (S \circ T) = \text{nullity} T$ from $\text{ker} S = \{0\}$.

For the second, I have no trouble showing $S \circ T$ is injective, but I do not understand how to show it is surjective.

Both questions are homework, so hints please, but I realise my issue is probably more with understanding how to find the dimenstion of an image.

Answer 1

With respect to your first question, note that

$$x\in \mathrm{Ker}(T\circ S)\iff (S\circ T)(x)=0\iff S(T(x))=0\underbrace{\iff}_{\mathrm{Ker}(S)=\{0\}} T(x)=0\iff x\in \mathrm{Ker}(T).$$

With respect to your second question, note that

$$z\in \mathbb{R}^n \underbrace{\implies}_{S\:\mathrm{onto}} \exists y\in \mathbb{R}^n: S(y)=z $$

$$y\in \mathbb{R}^n \underbrace{\implies}_{T\:\mathrm{onto}} \exists x\in \mathbb{R}^n: T(x)=y $$ from where

$$(S\circ T)(x)=S(T(x))=S(y)=z.$$

Answer 2

May this helps you:

Let $x\in ker(S\circ T)$

$\implies (S\circ T)(x)=0$

$\implies S(T(x))=0$

$\implies S(T(x))=S(0)$ since $S$ is a linear map $S(0)=0$

$\implies T(x)=0$ Since $S$ is injective

$\implies x\in kerT$

$\implies ker(S\circ T)\subseteq kerT$