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I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.

The question reads:

Let $V = F^{n}$. Establish a bijective correspondence between the sets $\mathscr{B}$ of bases of $V$ and $GL_{n}(F)$ (set of invertible matrices of size $n$ $x$ $n$ over Field $F$).

My proof goes as follows:

This seems rather straightforward once we appeal to the Change of Basis Formula. That is, $\forall{B} \in \mathscr{B}$ we can find a new basis $B'$ by plugging in

$B = PB'$ where P is an invertible matrix. This comes from the fact that, given sets $U$ and $S$, we know the elements of $U$ are in the span of $S$ if and only if there exists an $m$ $x$ $n$ matrix P such that

$(v_{1}, ..., v_{m})P = (w_{1},...,w_{n})$

We know P is invertible because, interchanging $B$ and $B'$ we know that there exists a matrix $P'$ such that $B' = P'B$. Substituting this in the original change of basis formula gets us

$BP'P = B$. That means

$(v_{1}, ..., v_{m})P'P = (v_{1},...,v_{n})$

Which means $(v_{1})P'_{(1)}P_{(1)} = v_{1}$

But, since $B$ is linearly independent (given that it's a basis) there is only one way for this to hold which is $B\textit{I} = B$. This implies $PP' = \textit{I}$ which means $P' = P^{-1}$

So, because P is invertible, there exists a one-to-one correspondence between every basis $B$ and every invertible matrix $P$ which can be accomplished by just rearranging the change of basis formula.

The one caveat to note is that $P$ must be unique. That is for every $B \in \mathscr{B}$ there is one matrix $P$ for which this holds. This can be proven a number of ways (probably most simply by appealing to the uniqueness of inverses, and therefore, since P is invertible, it must be some other element's unique inverse). However, more straightforwardly, this can be shown by contradiction. That is, assume there's some $A$ such that $B = AB'$ and $B = PB'$. That means that for some element $a_{ij}$ as an entry in A and $p_{ij}$ as an entry in P, $a_{ij} \neq p_{ij}$. However, we're just doing matrix matrix multiplication and therefore you won't get the same element in $B$

Is this correct? Thanks for your help!

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  • $\begingroup$ The change of basis formula is $B=B′P$, not $B=PB′$. Similarly, it's $B′=BP′$, not $B′=P′B$. Matrix multiplication is not commutative. $\endgroup$ – hchar Aug 30 '18 at 18:16

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