I'm a beginner in linear algebra, but i recently stumbled upon what seems to be a paradox to me.
Let $E$ be a $\mathbb{K}$-vector space with finite dimension ($\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$), $n=\dim(E)$, and $(e_i)_{1\le i\le n}$ be a basis of $E$.
We know that $(\mathcal{L}(E), +, .)$, the space of endomorphisms of $E$, is a vector space. Same with $(GL(E), +, .)$, the space of automorphisms of $E$.
We can prove that a basis of $\mathcal{L}(E)$ is the set of vectors $\mathcal{B}=(f_{i,j})_{(i,j)\in[1..n]\times[1..n]}$, where $\forall k\in[1..n], f_{i,j}(e_k)=\delta_{i,k}e_j$ and thus $\dim(\mathcal{L}(E))=n^2$.
We can also prove that there exists an isomorphism between $GL(E)$ and $\mathcal{M}_n(\mathbb{K})$, the space of square matrices with coefficients in $\mathbb{K}$. Which means $\dim(GL(E))=n^2=\dim(\mathcal{L}(E))$.
As far as i understand matrices yet, the matrix representation of $\mathcal{B}$ is also a basis of $\mathcal{M}_n(\mathbb{K})$.
Here comes what i believe to be a contradiction, and i hope someone can share clarification : how can $GL(E)\subsetneq\mathcal{L}(E)$, and $\mathcal{B}$ be both a basis of $GL(E)$ and $\mathcal{M}_n(\mathbb{K})$ ?
Thanks in advance.