# Is a choice of basis just a choice of isomorphism $M_n(\Bbb F)\to \text{End}_{\Bbb F}(\Bbb F^n)$?

I'm trying to learn more advanced mathematics, and I have a question. My brother has helped me type set this in $$\LaTeX$$, hopefully it's an okay question.

A vector space $$V$$ over a field $$\Bbb F$$ of dimension $$n$$ has an $$\Bbb F$$-algebra of linear endomorphisms $$\text{End}_{\Bbb F}(V)\cong \text{End}_{\Bbb F}(\Bbb F^n).$$ A linear endomorphism of $$V$$ can be identified with an $$n\times n$$ matrix, with coefficients in $$\Bbb F$$. That is to say, there is a set $$M_n(\Bbb F)$$ of matrices that, a priori, has nothing to do with endomorphisms of $$V$$. But, after a 'choice of basis' these matrices have a meaning. By this, of course, we mean that we define $$A\in M_n(\Bbb F)$$ as acting on $$v\in V$$ by: $$Mv=(m_{ij})_{i,j=1}^n(v_1,\dots,v_n)=\left(\sum_{k=1}^n m_{1k}v_k,\dots,\sum_{k=1}^nm_{nk}v_k\right).$$

My question is the following:

Is the choice of a basis nothing other than the choice of an isomorphism $$\text{End}_{\Bbb F}(\Bbb F^n)\cong M_n(\Bbb F)$$.

I apologise if this is completely obvious!

• No. For $n = 1$, there are $\left|\mathbb F\right|-1$ many bases, but only one isomorphism (assuming that you mean $\mathbb F$-algebra isomorphisms). – darij grinberg Nov 27 '19 at 6:38
• What is true is that the choice of a basis is "nothing other than" (i.e., in bijection with) the choice of an isomorphism of $\mathbb F$-modules $\mathbb F^n \to V$. – darij grinberg Nov 27 '19 at 6:39

That's pretty close; a basis is the choice of an isomorphism from $$\Bbb F^n$$ to $$V$$.
The vector $$(1, 0, 0, \ldots, 0)$$ goes to your first basis vector, the vector $$(0, 1, 0, 0, \ldots, 0)$$ goes to your second one, etc.
That isomorphism naturally induces an isomorphism between $$\mathrm{End}_{\Bbb F}(V)$$ with $$\mathrm{End}_{\Bbb F}(\Bbb F^n)$$. The latter can be canonically identified with $$n \times n$$ matrices.