$\mathbb{R}(n)$ is a simple algebra

Question

Let $\mathbb{R}(n)$ be the set of n by n real matrices.
An algebra $\mathcal{A}$ is said to be simple if (Lang pag 653):

1. $\mathcal{A}=\bigoplus_{i=1}^n I_i$ $\quad$ with $I_i$ being simple left ideals
2. there is exactly one class of isomorphism of simple left ideals.

I'm studying the book "Spin Geometry" by Lawson and Michelson and they say (with no proof) that $\mathbb{R}(n)$ is a simple algebra.
I already know that $$\mathbb{R}(n)=\bigoplus_{i=1}^n C_i$$ with $C_i$ being the subspace which have all the entries zero except possibly those in the i-th column, and i know that $C_i$ are simple left ideals isomorphic to each other.

My problem is: how to prove that ALL simple left ideals are isomorphic to each other?

Answer 1

Let I be a simple left ideal. Let $x \in I$ a non zero element and $f: \mathbb{R}(n) \to I$ such that $f(a)=ax$.
Now f is a non zero omomorphism so there exist an $i \in \{1,...,n\}$ such that $f|_{C_i}$ is non zero. But $f|_{C_i}$ is a non zero omomorphism between simple ideals, so it must be an isomorphism.
Now the thesis follows from the fact that those $C_i$ are isomorphic to each other.