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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!

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  • $\begingroup$ 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). $\endgroup$ – darij grinberg Nov 27 '19 at 6:38
  • $\begingroup$ 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 $\mathbb F^n \to V$. $\endgroup$ – darij grinberg Nov 27 '19 at 6:39
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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.

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