Proving a vector space cannot exist Hamilton tried to find a $3$-dimensional number system with the following properties:

*

*Every number can be written by $a + bx + cy$. This means every real number $a$ can be represented by $a + 0x + 0y$.


*Addition in this vector space satisfies commutativity, associativity, zero vectors, and additive inverses.


*If one number is real (i.e., $b = c = 0$), then multiplication is just standard scalar multiplication.


*Multiplication is associative.


*If one of the numbers is real, then multiplication is commutative. But in general, it is not commutative.


*Every nonzero element has a multiplicative inverse.

However, we will show that such a system cannot exist.  Suppose such a system did exist. Denote these numbers by $S$. We want to show $S$ cannot exist.
(a) Take any element $s \in S$ that is not real (so both $b$ and $c$ are not zero). Define the map $f : S \mapsto S$ by $f(v) = sv$. Show that $f$ is linear.
Since $f$ is linear, there is a $3\times 3$ matrix representing it. The characteristic polynomial of the matrix has degree $3$, so there is a real eigenvalue of $f$.
(c) The function $f$ says $f(w) = sw$ and since $w$ is an eigenvector, this implies $f(w) = \lambda v$, so we must have $\lambda v = sv$. But this implies $(\lambda - s)v = 0$. Which rule of multiplication is needed here to show $\lambda = s$?
(d) Why does $\lambda = s$ lead to a contradiction?

For (a), I need to show $f(a + b) = f(a) + f(b)$ and $f(\alpha a) = \alpha f(a)$, but I'm having trouble doing so. I know that to show the addition property, I need to use distributive properties and to show that it respects scalar multiplication, I need to utilize associativity of multiplication and the fact that the real numbers are commutative.
I tried taking one vector $v_1 = a_1 + b_1x + c_1y$ and $v_2 = a_2 + b_2x + c_2y$ so that
$$f(\alpha v_1) = f(\alpha a_1 + \alpha b_1x + \alpha c_1 y) = (a + bx + cy)(\alpha(a_1 + b_1x + c_1y)) = \alpha(a + bx + cy)(a_1 + b_1x + c_1y) $$
Is this right? I'm not really sure how to approach additivity.
(c) I'm not really sure what rule of multiplication is being used here. I think we're using the fact that every nonzero element has a multiplicative inverse so that we can multiply by the inverse of $v$ on both sides. Is this right?
(d) Because $\lambda$ is real and $s$ isn't

The proof can be synthesized in the following way. Let $S$ be a division algebra satisfying our conditions.

*

*In a division algebra for $a\neq 0$ and $b,c \in S$ we have that $ab=ac \implies b=c$  (cancellation).


*Given a non-real element $s$ multiplication on the right by $s$ must coincide with multiplication by a matrix $A$ of size $3$, so there is an eigenvalue $\lambda$ of $A$ with eigenvector $w$. It follows $w\lambda=ws$. Using cancellation contradicts the fact that $s$ is non-real (because $s=\lambda$).
