User:IssaRice/Linear algebra/Invertible equals expressible as change of coordinate matrix

Let $T:\mathbf {F} ^{n}\to \mathbf {F} ^{n}$ be invertible. We want to write $T$ as a change of coordinate matrix, i.e. as $[I]_{\beta }^{\beta '}$ for some appropriate choice of bases $\beta ,\beta '$. More precisely we want to write $[T]_{\sigma }^{\sigma }=[I]_{\beta }^{\beta '}$ where $\sigma =(e_{1},\ldots ,e_{n})$ is the standard basis of $\mathbf {F} ^{n}$. If we pick $\beta :=(Te_{1},\ldots ,Te_{n})$ and $\beta ':=\sigma$, then the kth column of $[I]_{\beta }^{\beta '}$ is $[ITe_{k}]^{\sigma }=[Te_{k}]^{\sigma }$, which is the kth column of $[T]_{\sigma }^{\sigma }$. Where did we use invertibility of $T$? To show that $(Te_{1},\ldots ,Te_{n})$ is a basis requires this; see User:IssaRice/Linear algebra/Properties of a list of vectors and their images.

The matrix $[I]_{\beta }^{\beta '}$ is invertible and its inverse is $[I]_{\beta '}^{\beta }$.

More generally, let $V$ be a finite-dimensional vector space and let $T:V\to V$ be an invertible linear transformation. Can we write $T$ as a change of coordinate matrix? One problem is that the question doesn't quite make sense. If our vectors aren't in $\mathbf {F} ^{n}$, then what does it mean to multiply a matrix by a vector? In the case of $V=\mathbf {F} ^{n}$ we could use the fact that $v=[v]^{\sigma }$.

This suggests that we can just pick a basis, say $\alpha$, and use it as our "standard" basis. Now we have $[T]_{\alpha }^{\alpha }=[I]_{(Tv_{1},\ldots ,Tv_{n})}^{\alpha }$.

So invertible linear maps are exactly those linear maps that "look like" the identity map, as long as we get to pick one of the bases.

Recall that there is a unique linear map that sends a basis to some list of vectors, and that the list of image vectors is a basis iff(?) the map is invertible. How can we interpret the result of this page in this context? If we start out with the standard basis $(e_{1},\ldots ,e_{n})$ then $T$ sends this to the basis $(Te_{1},\ldots ,Te_{n})$. But now the result says if we start with the basis $(Te_{1},\ldots ,Te_{n})$ and, without doing anything, just change our perspective to look at these vectors in terms of $(e_{1},\ldots ,e_{n})$, then this "perspective change operation" is identical to applying $T$. This sounds crazy, and I can't visualize it!