Representation Theorem for Finite Dimensional Algebra over Fields

Question

"Every finite dimensional algebra over a field has a faithful matrix representation."

Is the above statement true? If it is, how do I compute such a representation for the geometric algebra $\mathbb{G}^n$?

The question arises here Equivalence of Left and Right Inverse in Geometric Algebra where the statement is used to justify equivalence of left and right inverses.

if your algebra has a unit, the regular representation is faithful. — Mariano Suárez-Álvarez, May 14 '17 at 19:42
Please assume I am very bad at algebra. I mostly know about groups and rings, and almost nothing about representation theory. — Mark, May 14 '17 at 19:46
My comment has absolutely nothing to do with representation theory, really... — Mariano Suárez-Álvarez, May 14 '17 at 19:59

score 3 · Accepted Answer · answered May 14 '17 at 19:59

3

Let $A$ be a finite dimensional algebra with $1$ over a field $k$, let $n$ be it dimension and let $\mathcal B$ be an ordered basis of $A$. For each $a\in A$ there is a linear map $r_a:x\in A\mapsto ax\in A$ given by multiplication on the left by $a$. Let $R(a)$ be the matrix of $r_a$ with respect to the basis $\mathcal B$.

The function $R:a\in A\mapsto R(a)\in M_n(k)$ is a faithful matrix representation of $A$.

answered May 14 '17 at 19:59

Mariano Suárez-Álvarez

139,300

Why must it be faithful? Could there be two vectors that both act as identity? – Mark May 14 '17 at 20:06
Oh then just multiply them together to show they are equal – Mark May 14 '17 at 20:12
3

Hm? I don't know what you mean by 'multiply them together"... In any case: the function $R$ is injective simply because when $a$ and $b$ are different elements of $a$ the values $r_a(1)$ and $r_b(1)$ are different. – Mariano Suárez-Álvarez May 14 '17 at 20:24
"Let $R(a)$ be the matrix of $r_a$" already presupposes that there exists a matrix representation. – user48672 May 22 '17 at 13:39
@user48672, huh? No: I defined a linear map $r_a$ and I am taking the matrix of that linear map with respecto a fixed ordered basis. – Mariano Suárez-Álvarez May 22 '17 at 15:30

score 0 · Answer 2 · answered May 14 '17 at 19:42

0

This result is true if you suppose that the algebra is simple.

https://en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem

answered May 14 '17 at 19:42

Tsemo Aristide

89,587

5

And also if you don't suppose that. – Mariano Suárez-Álvarez May 14 '17 at 20:01

Representation Theorem for Finite Dimensional Algebra over Fields

2 Answers2

Linked