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

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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!