Let $Q$ be an invertible $n\times n$ matrix, $A$ an $n\times n$ matrix and $x$ is an $n\times 1$ column vector (or matrix) so that the matrix equation $Ax = 0$ represents a homogeneous system of $n$ equations in $n$ unknowns. Show that if $(QA)x = 0$ has just the trivial solution, then $A$ is invertible.
All I can think of is letting $Q$ be equal to $A^{-1}$ since its invertible, then multiplying them to get $I$. then if $Ix = 0$ is only trivial this proves $Q$ can be $A^{-1}$. But I'm pretty sure i have this in the wrong order with the wrong wording, or might be wrong all together.. can anyone help me with this? I am bad at formal proofs.
Best Answer
For the direction you want to prove, the invertibility of $Q$ is a distraction: suppose that $A$ is not invertible, then $A$ doesn't have full column rank so that there is $x\neq 0$ such that $Ax=0$. Then $(QA)x=Q(Ax)=Q\times 0=0.$
You only need the invertibility of $Q$ if you want to prove the reverse direction: suppose that $A$ is invertible and suppose that $x\neq 0$, then $Ax\neq 0$. Then, because $Q$ is invertible, you have $(QA)x=Q(Ax)\neq 0$. This means that if $A$ is invertible, then $(QA)x=0$ only has the trivial solution.