# Abstract Nonsense

## Algebra of Square Matrices (Pt. II)

Point of post: This post is a continuation of this post. We will finish the proof started in the last post and then wrap up our discussion of square matrices as independent algebraic objects.

(continuing the proof at the end of the last post) We now prove that the distributivity property holds. To do this let $M,N,$ and $P$ be as above and note that, by definition, the general entry of $M(N+P)$ is

$\displaystyle \sum_{r=1}^{n}\alpha_{i,r}\left(\beta_{r,j}+\gamma_{r,j}\right)=\sum_{r=1}^{n}\alpha_{i,r}\beta_{r,j}+\sum_{r=1}^{n}\alpha_{i,r}\gamma_{r,j}$

which is the general term for $MN+MP$. The other distributivity axiom follows similarly.

To prove the scalar distributivity property we merely note that if $M$ and $N$ are as before and $\alpha\in F$ then, by definition the general entry of $\alpha(MN)$ is

$\displaystyle \alpha \sum_{r=1}^{n}\alpha_{i,r}\beta_{r,j}=\sum_{r=1}^{n}(\alpha \alpha_{i,r})\beta_{r,j}=\sum_{r=1}^{n}\alpha_{i,r}(\alpha \beta_{r,j})$

But, these last two forms are the general entry of $(\alpha M)N$ and $M(\alpha N)$ respectively.

It remains to show that $I_n$ is, in fact, a multiplicative identity. To do this it suffices to note that if $M$ is as before then the general entry of $MI_n$ is

$\displaystyle \sum_{r=1}^{n}\alpha_{i,r}\delta_{r,j}=\alpha_{i,j}\delta_{j,j}=\alpha_{i,j}$

which is the general entry for $M$. Similarly, the general entry for $I_nM$ is

$\displaystyle \sum_{r=1}^{n}\delta_{i,r}\alpha_{r,j}=\delta_{i,i}\alpha_{i,j}=\alpha_{i,j}$

which is the general entry for $M$. It follows that $MI_n=I_nM=M$ as required.

Thus, since all the axioms have been verified 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