Invertibility of the operands of an invertible product of matrices

Question

Let $S$ and $T$ be two matrices. Assume that $ST$ is invertible. I want to prove that $S$ and $T$ are invertible.

I managed to prove that $T$ is invertible. Here is my proof. Assume $T(X)=0$. Then $X=I X=(ST)^{-1}(ST)X=(ST)^{-1}S(0)=0$. So $T$ is invertible.

How to prove that $S$ is also invertible ?

Answer 1

I am assuming that $s$ and $T$ are square matrices. To see that $S$ is invertible, assume that $x$ is a vector of appropriate size and that $xS=0$. Then by your previous argument, $x=xI=x(ST)(ST)^{-1}=(0)T(ST)^{-1}=0$.

Answer 2

$$(ST)^{-1} = T^{-1}S^{-1}$$

Therefore, if $(ST)^{-1}$ exists and $S,T$ are square, then they are invertible as well.

Answer 3

Since $ST$ is invertible it must be a bijection. This in particular must mean that $S$ is surjective and $T$ is injective.

Injective maps are always left-invertible and surjective maps are always right-invertible. If $S$ and $T$ are square, then the two are equivalent and both are simply invertible.

To be more explicit, let $U$ be the inverse of $ST$. Then $$U(ST) = (US)T = I$$ so $T$ is left invertible with inverse $US$ and $$(ST)U = S(TU) = I$$ so that $S$ is right invertible with inverse $TU$.

Answer 4

If either $S$ or $T$ is not invertible, then $\det(ST)=\det(S)\det(T)=0$, so ...