What does $GL(V) \cong GL_n(\mathbb{C})$ mean? I have proven that $GL(V) \cong GL_n(\mathbb{C})$, where $GL(V)=\{f: V \rightarrow V | \text{ f is bijectiv and linear}\}$ and $GL_n(\mathbb{C})$ is the set of $n \times n$ complex matrices with non-zero determinant and with the operation multiplication given as multiplication of matrices. 
Why is it important to show that $GL(V) \cong GL_n(\mathbb{C})$? What do we use this for? I know im not specific in my questions. But why do we want to show this?  
 A: $GL(V)$ is much harder to visualise than $GL_n(\mathbb{C})$, and the isomorphism says they are the same. So you can work with 'easy objects' instead. In this concrete example, this means that you fix a basis and do the computations w.r.t. this basis.
A: The elements of $GL(V)$ are linear maps, in particular applications $V\to V$. They can be quite complicated a priori. 
The elements of $GL_n(\mathbb C)$ are matrices, so literally tables of numbers : those seem very easy, and computations are more likely to be tractable with them than with linear maps. 
On the other hand, $V$ has in general no preferred basis, so reasoning with $GL(V)$ can be seen as working "coordinate-free"; and can sometimes be more conceptual than some raw matrix computations, so it can also have its benefits. 
The isomorphism you note allows for a dictionary between the two approaches; if you need to compute stuff, pick a basis, an isomorphism, and compute with matrices. If you need to do some conceptual work, forget the bases, and work "geometrically" with $V$. 
A: Here is an interesting example.
Consider $P_n$, of polynomials with complex coefficients of degree $n-1$. $P_n$ is a $n$ dimensional vector space over $\mathbb{C}$. The derivative map $f(x) \mapsto f'(x)$ is a linear operator on $P_n$. 
The theorem shows that there is some $n\times n$ matrix $M$ with the following property: for all $c_1, ..., c_n, d_1, ..., d_n$: $$M \times [c_1, ..., c_n ]^T = [d_1, ..., d_n ]^T \iff \frac{d}{dx}\{c_1 + c_2 x + c_3 x^2 + ... + c_n x^{n-1}\} = \{d_1 + d_2 x + d_3 x^2 + ... + d_n x^{n-1}\}$$
In other words, taking a derivative is a specific form of matrix multiplication. More generally, all (finite dimensional) linear transformations are matrix multiplications.
