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