Show 3D-division algebra over the reals cannot exist using linear algebra

Question

There is a great comment by Jyrki Lahtonen here: Why is quaternion algebra 4d and not 3d?

It is not too difficult to show that a 3D-division algebra over the reals cannot exist. If $D$ were such a beast, and $x\in D$, $x \notin \mathbf{R}$, then consider the left regular representation of $D$ by 3x3 real matrices. The matrix $A$ representing multiplication by $x$ from the left cannot be scalar, because $x$ was not one. OTOH the eigenvalue polynomial of $A$ is cubic, and hence has a root $\lambda \in \mathbf{R}$. We have just shown that $x−\lambda$ is not invertible.

This sounds wonderful, but is just outside my understanding. It would be great if it is really possible to see the non-existence of a type of number system just using polynomials and linear algebra.

Can someone please fill in the details (even just link to wiki articles to read up on)? For instance:

1) Are we ruling out possible number systems by assuming there is a matrix representation?

Just by drawing a cubic polynomial, I can see one end must go to $+\infty$ and the other to $-\infty$, so it makes sense that there must always be at least one real root. I don't understand how this is useful though. How do we even know the Reals are a subset of these new numbers? (For instance with 3x3 matrices the the reals would be the 3x3 matrices which are just diagonals with all the same number, but maybe those aren't included in this representation of 3D numbers. After all, not all 2x2 matrices are complex numbers in the 2x2 matrix representation. Maybe for a 3D number system, each component must be "imaginary"?)

2) How does there being a real root prove there isn't a division algebra? I don't follow the argument there. How did polynomials come in here, and how did this show something wasn't invertible?

To help understand this better, how does this argument fail with the Reals, Complex numbers, or Quaternions? For example with the Reals, we could consider the trivial representation by 1x1 real matrices. These would have a degree one eigenvalue polynomial and therefore also have one real root.

Answer 1

1) no, we don't. If $D$ were a real division algebra of dimension $3$, then there is an underlying structure of a real vector space and left multiplication by $x$ is just a real linear endomorphism. These are known to have this kind of representation (by real 3x3 matrices). If you don't know this you will need to learn some more linear algebra. Moreover, as $D$ is supposed to be free of zero divisors, it must be even an isomorphism.

Adressing your other questions, you need to know that the eigenvalues of a real endomorphism are given by the roots of it's characteristic polynomial, which, for the three dimensional case happens to be a polynomial of degree 3. This is just basic linear algebra, and again, if you did not come to know this by now, you will not be able to understand the reasoning here. This fact 'is useful', to directly address your concern, because it implies (as stated in your citation) that the endomorphism $x-\lambda id$ cannot have full rank. Now going back again to the structure of a division algebra, this element $x-\lambda id$ is now zero divisor, different from $0$. This is just plugging in the defnitions.

In the 2d and 4d case you cannot prove that the characteristic polynomial has (real) a zero, because it will have degree 2 or 4. A shifted parabola,e.g, will have no real zero. In the one d-case it is not possible to find an $x$ which is not a real (not being a real is an assumption for $x$ the proof starts out with).

Note added: it is interesting to note that Sir William Rowan Hamilton, a brillant mathematician, tried for his whole lifetime to find a three dimensional number system (in vain of course, since they cannot exist). If you look at the proof you reproduced here then you might wonder how this is possible. One answer is that a lot of mathematical research and knowledge is coded deeply in the definitions and structures we are using today.