# 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!