## General Linear Groups — Generally

Monday, we saw that the general linear groups are matrix groups, specifically consisting of those whose columns are linearly independent. But what about more general vector spaces?

Well, we know that every finite-dimensional vector space has a basis, and is thus isomorphic to , where is the cardinality of the basis. So given a vector space with a basis of cardinality , we have the isomorphism defined by and .

This isomorphism of vector spaces then induces an isomorphism of their automorphism groups. That is, . Given an invertible linear transformation , we can conjugate it by to get . This has inverse , and so is an element of . Thus (not unexpectedly) every invertible linear transformation from a vector space to itself gets an invertible matrix.

But this assignment depends essentially on the *arbitrary choice* of the basis for . What if we choose a different basis ? Then we get a new transformation and a new isomorphism of groups . But this gives us an inner automorphism of . Given a transformation , we get the transformation

This composite sends to itself, and it has an inverse. Thus changing the basis on induces an inner automorphism of the matrix group .

Now let’s consider a linear transformation . We have two bases for , and thus two different matrices — two different elements of — corresponding to : and . We get from one to the other by conjugation with :

And what is this transformation ? How does it act on a basis vector in ? We calculate:

where expresses the vectors in one basis for in terms of those of the other. That is, the th column of the matrix consists of the components of written in terms of the . Similarly, the inverse matrix with entries , writes the in terms of the : .

It is these “change-of-basis” matrices that effect all of our, well, changes of basis. For example, say we have a vector with components . Then we can expand this:

So our components in the new basis are .

As another example, say that we have a linear transformation with matrix components with respect to the basis . That is, . Then we can calculate:

and we have the new matrix components .

[...] we’re drawn again to consider the special case where . Now an isomorphism is just a change of basis. Representations of are equivalent if they do “the same thing” to the vector space , [...]

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