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 $A$?
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:

$A\Biggl(\begin{matrix} x\_1 \\ ... \\ x\_n \end{matrix}\Biggr)=\Biggl(\begin{matrix} 0 \\ ... \\ 0 \end{matrix}\Biggr)$ then $x\_1,...,x\_n = 0$, 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.