Question: We have $A\in \operatorname{Mat}\_n(K)$ and I am told to prove that $A\in GL\_n(K)$ if and only if meets
$A\Biggl(\begin{matrix} x\_1 \\ ... \\ x\_n \end{matrix}\Biggr)=\Biggl(\begin{matrix} 0 \\ ... \\ 0 \end{matrix}\Biggr)$ 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 $f\in \operatorname{End}V)$ whose associated matrix is $A$.

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

Answer: 
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 $x=0$.