T and S are invertible and why (ST) is invertible too? I have functions T and S such that T : V -> V  and S : V->V 
and S and T are invertible.
Prove why composition of (ST) is invertible too.
I tried to figure it out by letting TS=ST, and TS and ST are linear.
but I am stuck... 
 A: Hint: A map (on a finite dimensional vector space) is invertible if and only if it has trivial nullspace, i.e. if $Ax = 0$, then $x = 0$. What can you say about the nullspace of $ST$ with respect to the nullspaces of $S$ and $T$?
To show that $(ST)^{-1} = T^{-1}S^{-1}$, try multiplying $ST$ and $T^{-1}S^{-1}$. What do you get?
A: Set Theoretic Approach. This can be generalized to $T: V \to W$ and $g : W \to U.$ 
Injectivity:
\begin{align}
STx = STy &\implies Tx = Ty \quad \text{because $S$ is injective.}\\
&\implies x = y \quad \text{because $T$ is injective.}
\end{align}
Surjectivity:
Suppose $v \in V$; since $S$ is surjective, $\exists b \in V$ such that $v = Sb$. As $b \in V$, then by the surjectivity of $T$, $\exists a \in V$ such that $b = Ta$. Hence we immediately see that for any $v \in V$, there exists $a \in V$ such that $v = Sb = STa$. 
Determinant:
There is also a way to see this through determinants, that is $$0 \neq c = \det T \det S = \det TS .$$
A: The proof has two distinct aspects.


*

*If $f:A\to B$ and $g:B\to C$ are any invertible (= bijective) maps between sets, then their composition $g\circ f:A\to C$ is always also invertible. The inverse is the composition of $\def\'{^{-1}}g\':C\to B$ and $f\':B\to A$, in other words $(g\circ f)\'=f\'\circ g\'$. This part is completely general and has no specific relation to linear algebra.

*If a map between vector spaces $T:V\to W$ is both linear and bijective (invertible as a map of sets), then the inverse map of sets $f\':W\to V$ is also linear. This is a pure linear algebra fact that should be one of the first things to know about invertible linear maps. It's proof looks a bit tricky, but it you think of it is quite natural (use the linearity of $f$ in the opposite sense, after inserting back-and-forth maps to make $f$ appear in the first place):
$$
  f\'(\lambda w_1+\mu w_2)
 =f\'\bigl(\lambda f(f\'(w_1))+\mu f(f\'(w_2))\bigr)
 =f\'\bigl( f\bigl(\lambda f\'(w_1)+\mu f\'(w_2)\bigr)\bigr)
 =\lambda f\'(w_1)+\mu f\'(w_2).
$$
Now all that is left is apply it to your situation: $ST$ in a bijective map (1) with inverse $T\'S\'$, and (2) this inverse is a linear map (since you checked that $ST$ is linear).
