The Unapologetic Mathematician

Mathematics for the interested outsider

General Linear Groups — Generally

Monday, we saw that the general linear groups \mathrm{GL}_n(\mathbb{F}) 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 \mathbb{F}^n, where n is the cardinality of the basis. So given a vector space V with a basis \{f_i\} of cardinality n, we have the isomorphism S:\mathbb{F}^n\rightarrow V defined by S(e_i)=f_i and S^{-1}(f_i)=e_i.

This isomorphism of vector spaces then induces an isomorphism of their automorphism groups. That is, \mathrm{GL}(V)\cong\mathrm{GL}_n(\mathbb{F}). Given an invertible linear transformation T:V\rightarrow V, we can conjugate it by S to get S^{-1}TS:\mathbb{F}^n\rightarrow\mathbb{F}^n. This has inverse S^{-1}T^{-1}S, and so is an element of \mathrm{GL}_n(\mathbb{F}). Thus (not unexpectedly) every invertible linear transformation from a vector space V to itself gets an invertible matrix.

But this assignment depends essentially on the arbitrary choice of the basis \{f_i\} for V. What if we choose a different basis \{\tilde{f}_i\}? Then we get a new transformation \tilde{S} and a new isomorphism of groups T\mapsto\tilde{S}^{-1}T\tilde{S}. But this gives us an inner automorphism of \mathrm{GL}_n(\mathbb{F}). Given a transformation M:\mathbb{F}^n\rightarrow\mathbb{F}^n, we get the transformation
This composite \tilde{S}^{-1}S sends \mathbb{F}^n to itself, and it has an inverse. Thus changing the basis on V induces an inner automorphism of the matrix group \mathrm{GL}_n(\mathbb{F}).

Now let’s consider a linear transformation T:V\rightarrow V. We have two bases for V, and thus two different matrices — two different elements of \mathrm{GL}_n(\mathbb{F}) — corresponding to T: S^{-1}TS and \tilde{S}^{-1}T\tilde{S}. We get from one to the other by conjugation with \tilde{S}^{-1}S:


And what is this transformation \tilde{S}^{-1}S? How does it act on a basis vector in \mathbb{F}^n? We calculate:
where f_j=x_j^i\tilde{f}_i expresses the vectors in one basis for V in terms of those of the other. That is, the jth column of the matrix X consists of the components of f_j written in terms of the \tilde{f}_i. Similarly, the inverse matrix X^{-1} with entries \tilde{x}_i^j, writes the \tilde{f}_j in terms of the f_i: \tilde{f}_i=\tilde{x}_i^jf_j.

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


So our components in the new basis are \tilde{v}^i=x_k^iv^k.

As another example, say that we have a linear transformation T:V\rightarrow V with matrix components t_i^j with respect to the basis \{f_i\}. That is, T(f_i)=t_i^jf_j. Then we can calculate:


and we have the new matrix components \tilde{t}_i^j=\tilde{x}_i^kt_k^lx_l^j.

October 22, 2008 - Posted by | Algebra, Group Examples, Linear Algebra


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