# Abstract Nonsense

## 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 $\mathscr{V}$ be an $n$-dimensional $F$-space. Then, we define the $n$-tuple $\mathcal{B}=(x_1,\cdots,x_n)$ to be an ordered basis for $\mathscr{V}$ if $\{x_1,\cdots,x_n\}$ is a basis for $\mathscr{V}$. 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 $\mathscr{V}$ was as in the last paragraph and let $T\in\text{End}\left(\mathscr{V}\right)$. Then, if $\mathcal{B}$ is an ordered basis for $\mathscr{V}$  we may define the matrix $\left[T\right]_{\mathcal{B}}$ as follows: for each $j\in[n]$ we know that there exists scalars $\alpha_{1,j},\cdots,\alpha_{n,j}\in F$ such that

$\displaystyle T(x_j)=\sum_{i=1}^{n}\alpha_{i,j}x_i$

Thus, doing this for $i=1,\cdots,n$ we arrive at $n^2$ scalars $\left\{\alpha_{i,j}:i,j\in[n]\right\}$. We then define

$\left[T\right]_{\mathcal{B}}=\begin{pmatrix}\alpha_{1,1} & \cdots & \alpha_{1,n}\\ \vdots & \ddots & \vdots\\ \alpha_{n,1} & \cdots & \alpha_{n,n}\end{pmatrix}$

To look at this differently we notice that given this ordered basis $\mathcal{B}$ we are afforded the canonical isomorphism between $\mathscr{V}$ and $F^n$ given by

$\displaystyle \phi_{\mathcal{B}}\left(\sum_{r=1}^{n}\alpha_r x_r\right)=\left(\alpha_1,\cdots,\alpha_n\right)$

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

$\displaystyle \phi_{\mathcal{B}}\left(\sum_{r=1}^{n}\alpha_r x_r\right)=\begin{pmatrix}\alpha_1\\ \vdots\\ \alpha_n\end{pmatrix}$

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

$\left[T\right]_{\mathcal{B}}=\left(\begin{array}{c|c|c} & & \\ \phi_{\mathcal{B}}\left(T(x_1)\right) & \cdots & \phi_{\mathcal{B}}\left(T(x_n)\right)\\ & & \end{array}\right)$

It is clear now why we needed to order the basis. In particular, if we had picked some other ordering of the basis $\{x_1,\cdots,x_n\}$ 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 $T\mapsto\left[T\right]_{\mathcal{B}}$ is, in fact, an associative unital algebra isomorphism.

Theorem: Let $\mathscr{V}$ be an $n$-dimensional $F$-space. Then, if $\mathcal{B}$ is an ordered basis for $\mathscr{V}$ the mapping

$[\cdot]_{\mathcal{B}}:\text{End}\left(\mathscr{V}\right)\to\text{Mat}_n\left(F\right):T\mapsto\left[T\right]_{\mathcal{B}}$

is an associative unital algebra isomorphism.

Proof: Let $T,T'\in\text{End}\left(\mathscr{V}\right)$ and $\alpha,\beta\in F$. Note then that if

$\displaystyle T(x_j)=\sum_{i=1}^{n}\alpha_{i,j}x_i$

and

$\displaystyle T'(x_j)=\sum_{i=1}^{n}\beta_{i,j}x_i$

then

\displaystyle \begin{aligned}\left(\alpha T+\beta T'\right)(x_j) &=\alpha T(x_j)+\beta T'(x_j)\\ &=\alpha\sum_{i=1}^{n}\alpha_{i,j}x_i+\beta\sum_{i=1}^{n}\alpha'_{i,j}x_i\\ &=\sum_{i=1}^{n}\left(\alpha \alpha_{i,j}+\beta \beta_{i,j}\right)x_i\end{aligned}

It follows then that the general entry of $\left[\alpha T+\beta T'\right]_{\mathcal{B}}$ is $\alpha \alpha_{i,j}+\beta \beta_{i,j}$ which is the same as the general entry of $\alpha\left[T\right]_{\mathcal{B}}+\beta\left[T'\right]_{\mathcal{B}}$. It follows that

$\left[\alpha T+\beta T'\right]_{\mathcal{B}}=\alpha\left[T\right]_{\mathcal{B}}+\beta\left[T'\right]_{\mathcal{B}}$

To see that $[\cdot]_{\mathcal{B}}$ preserves products let $T,T'$ be as before, except this time write

$\displaystyle T'(x_j)=\sum_{r=1}^{n}\beta_{r,j}x_r$

and

$\displaystyle T(x_j)=\sum_{i=1}^{n}\alpha_{i,j}x_i$

We then see that

\displaystyle \begin{aligned}\left(TT'\right)(x_j) &= T\left(T'(x_j)\right)\\ &= T\left(\sum_{r=1}^{n}\beta_{r,j}x_j\right)\\ &= \sum_{r=1}^{n}\beta_{r,j}T(x_j)\\ &= \sum_{r=1}^{n}\beta_{r,j}\sum_{i=1}^{n}\alpha_{i,r}x_i\\ &= \sum_{i=1}^{n}\left(\sum_{r=1}^{n}\alpha_{i,r}\beta_{r,j}\right)x_i\end{aligned}

so that the general term of $\left[TT'\right]_{\mathcal{B}}$ is $\displaystyle \sum_{r=1}^{n}\alpha_{i,r}\beta_{r,j}$ which is precisely the general term of $\left[T\right]_{\mathcal{B}}\left[T'\right]_{\mathcal{B}}$ from where it follows that

$\left[TT'\right]_{\mathcal{B}}=\left[T\right]_{\mathcal{B}}\left[T'\right]_{\mathcal{B}}$

Last but not least, noticing that for each $j\in[n]$ we have that

$\displaystyle \mathbf{1}(x_j)=\sum_{i=1}^{n}\delta_{i,j}x_i$

we see that the general term of $\left[\mathbf{1}\right]_{\mathcal{B}}$ is $\delta_{i,j}$ which is precisely the general term for $I_n$. Thus, $\left[\mathbf{1}\right]_{\mathcal{B}}=I_n$. It follows then that $[\cdot]_{\mathcal{B}}$ satisfies all the axioms for an associative unital algebra isomorphism.

Noticing then that $[\cdot]_{\mathcal{B}}$ is evidently bijective the conclusion follows.. $\blacksquare$

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