# Invertibility of the operands of an invertible product of matrices

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 ?

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$T$ is surjective. I am assuming that $S,T$ are square matrices. –  Will Jagy Oct 24 '12 at 21:11

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, $x=xI=x(ST)(ST)^{-1}=(0)T(ST)^{-1}=0$.

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Just like you can multiply a matrix on the right with a column vector, you can multiply it on the left with a row vector. To calculate it you can view it as $xA=(A^Tx^T)^T$ and do the multiplication like usual, since here $x^T$ is a column vector. –  espen180 Oct 24 '12 at 21:40
Thank you espen. –  Bob Oct 24 '12 at 21:42
My pleasure. :) –  espen180 Oct 24 '12 at 21:44

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

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

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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 $TU$.

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Great demo. Thank you ! –  Bob Oct 24 '12 at 21:36
If either $S$ or $T$ is not invertible, then $\det(ST)=\det(S)\det(T)=0$, so ...