User:IssaRice/Linear algebra/Invertible equals expressible as change of coordinate matrix
Let be invertible. We want to write as a change of coordinate matrix, i.e. as for some appropriate choice of bases . More precisely we want to write where is the standard basis of . If we pick and , then the kth column of is , which is the kth column of . Where did we use invertibility of ? To show that is a basis requires this; see User:IssaRice/Linear algebra/Properties of a list of vectors and their images.
The matrix is invertible and its inverse is .
More generally, let be a finite-dimensional vector space and let be an invertible linear transformation. Can we write as a change of coordinate matrix? One problem is that the question doesn't quite make sense. If our vectors aren't in , then what does it mean to multiply a matrix by a vector? In the case of we could use the fact that .
This suggests that we can just pick a basis, say , and use it as our "standard" basis. Now we have .
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 then sends this to the basis . But now the result says if we start with the basis and, without doing anything, just change our perspective to look at these vectors in terms of , then this "perspective change operation" is identical to applying . This sounds crazy, and I can't visualize it!