## Algebra Isomorphism Between Algebra of Square Matrices and the Endomorphism Algebra (Pt. II)

**Point of post: **This is a continuation of this post.

*Square Matrix Algebra and Endomorphism Algebra are Associative Unital Algebra Isomorphic *

In this section we discuss the big theorem in our discussion of matrices. Namely, we’ll show how to, given a fixed ordered basis (see below), we can associate with each endomorphism a matrix. Moreover, we’ll show that this association is an associative unital algebra isomorphism. To do this we begin by defining an “ordered basis”.

Let be an -dimensional -space. Then, we define the -tuple to be an *ordered basis *for if is a basis for . In other words, an ordered basis is just a basis for which we’ve designated which is the first, second, etc. basis vector.

Now, suppose that was as in the last paragraph and let . Then, if is an ordered basis for we may define the matrix as follows: for each we know that there exists scalars such that

Thus, doing this for we arrive at scalars . We then define

To look at this differently we notice that given this ordered basis we are afforded the canonical isomorphism between and given by

Or, recalling our convention about writing elements of as column vectors the above can be restated as

With this then we can see that our matrix (intuitively, this isn’t supposed to be rigorous) is really

It is clear now why we needed to order the basis. In particular, if we had picked some other ordering of the basis we would have arrived at a matrix whose set of entries was the same, but for which the rows are permuted.

We now claim that his association is, in fact, an associative unital algebra isomorphism.

**Theorem: ***Let be an -dimensional -space. Then, if is an ordered basis for the mapping*

*is an associative unital algebra isomorphism.*

**Proof: **Let and . Note then that if

and

then

It follows then that the general entry of is which is the same as the general entry of . It follows that

To see that preserves products let be as before, except this time write

and

We then see that

so that the general term of is which is precisely the general term of from where it follows that

Last but not least, noticing that for each we have that

we see that the general term of is which is precisely the general term for . Thus, . It follows then that satisfies all the axioms for an associative unital algebra isomorphism.

Noticing then that is evidently bijective the conclusion follows..

**References:**

1. Golan, Jonathan S. *The Linear Algebra a Beginning Graduate Student Ought to Know*. Dordrecht: Springer, 2007. Print.

2. Halmos, Paul R. *Finite-dimensional Vector Spaces,*. New York: Springer-Verlag, 1974. Print

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