proving a system with $A\vec{x} = \vec{0}=A^k\vec{x} $

Question

The question: "Let $A\vec{x} = \vec{0}$ be a homogeneous system of $n$ linear equations in $n$ unknowns that has only the trivial solution. Show that if $k$ is any positive integer, then the system $A^k\vec{x} = \vec{0}$ also has only the trivial solution." I attempted to prove it by saying that because $\vec{x}$ must be $\vec{0}$, then whatever matrix multiplied by it will be the $\vec{0}$ vector. However, I don't feel that this proves it well at all...

Answer 1

$Ax=0 \implies x=0$ means that $\det A \neq 0$, which means $\det A^k = (\det A)^k \neq 0$, so $A^k $ is nonsingular.

Answer 2

by induction

by hypothesis it is true for $k=1$

Assume $A^k x = 0$ has $x=0$ as the only solution

then $A^{k+1}x=0$ if and only if $A y = 0 $ where $y= A^k x$

but the only solution to $A y = 0 $ is $y=0 $ ( by the original hypothesis )

so the only solutions to $A^{k+1}x=0$ are the solutions to $y=A^k x = 0$

which by assumption possesses only the solution $x=0$

Therefore $A^{k+1}x=0$ possesses only the trivial solution and the theorem is proved .