# $QRx = 0$ is equivalent to $Rx = 0$?

I have read that if you are looking for the solution to $Ax = 0$ and perform a QR decomposition the following statement holds

$Ax = 0 \Longleftrightarrow QRx = 0 \Longleftrightarrow Rx = 0$

I can't see how $\Longleftrightarrow QRx = 0 \Longleftrightarrow Rx = 0$...the mapping $QR$ is not the same as $R$ so how can the kernel remain unchanged?

$Q$ is non-singular, such that $Q^{-1}$ exists and $$QRx=0 \iff Rx=Q^{-1}0 \iff Rx=0$$