A crucial concept in linear algebra is that the composition of two invertible linear transformations is itself invertible. Here is the first proof I learned of this fact:
Proof: Suppose that $T_1: \mathbb{C}^n \to \mathbb{C}^n$ and $T_2: \mathbb{C}^n \to \mathbb{C}^n$ are both invertible with respective matrices $A_1$ and $A_2$. Then the matrix of their composition $T_2 \circ T_1$ is simply $A_2A_1$. Since $T_1$ and $T_2$ are invertible, we know that $\det(A_1) \neq 0$ and $\det(A_2) \neq 0$. Thus, we see that $\det(A_1A_2) = \det(A_1)\det(A_2) \neq 0$. Thus, the composition is also invertible. $\square$
I'm trying to now give a proof of this fact without using determinants. Any idea on where to start?
Best Answer
In finite dimensional space it suffices to prove that $T_2\circ T_1$ is injective:
$$ x\in\ker(T_2\circ T_1)\iff T_2(T_1(x))=0\iff T_1(x)=0\iff x=0 $$ We used in the second equivalence the fact that $T_2$ is injective and in the last equivalence the fact that $T_1$ is injective. Conclude.