Vrftjkry

We have $Ain operatorname{Mat}_n(K)$ and I am told to prove that $Ain GL_n(K)$ if and only if meets
$ABiggl(begin{matrix} x_1 \ ... \ x_n end{matrix}Biggr)=Biggl(begin{matrix} 0 \ ... \ 0 end{matrix}Biggr)$ then $x_1,...,x_n = 0$

As a tip I am told to consider $beta$ the base of a vector space $V$ with $n$ dimension and $fin operatorname{End}V)$ whose associated matrix is $A$.

I don't know how to use that "tip" to prove that.
Thanks in advance.

As indicated by your tip, consider the endomorphism $f$ represented by this matrix in a given basis of a $K$-vector space $V$ with dimension $n$.

Remember that $A$ is invertible if and only if $f$ is an isomorphism (more exactly, an automorphism since $f$ is an endomorphism).

Now in a finite dimensional space, $f$ is an automorphism if and only if it is injective (and also if and only if it is surjective).

Now, how do you characterise an injective linear map?

Some more details:

$f$ is injective (hence bijective) if and only if $ker f={0}$, i.e. if and only if
$$(f(v)=0)implies (v=0).$$
What is the relation between $f(v)$ and $A$?

To get you started: A matrix operation is invertible if it is injective and surjective. If $Ax=0$ for some non-zero vector $x$, then $A$ is not injective, so it is not invertible. The contrapositive is that $A$ being invertible implies $Ax=0$ only for $x=0$.

If $A$ is invertible, then using $Rank-nullity$ $theorem$, we can proove that $text{Ker}(A) = {0}$, so if $A(x) = 0$ then $x=0$.

Reciprocally, if $A$ meets the condition:

$ABiggl(begin{matrix} x_1 \ ... \ x_n end{matrix}Biggr)=Biggl(begin{matrix} 0 \ ... \ 0 end{matrix}Biggr)$ then $x_1,...,x_n = 0$, it means that $text{Ker}(A) = {0}$. Using again the $Rank-nullity$ $theorem$, we can prove that $text{Rank}(A) = n$, which ensures us that $A$ is invertible.