# 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

• This all looks good. For additivity in $a$ you just have to use the distributive property of the number system. – Yorch May 7 at 13:22
• You technically did not require any distributivity condition between your addition and multiplication, which I guess makes most of the proofs incorrect. But anyway what you want to prove is that "there is no 3-dimensional division algebra over $\mathbb{R}$". – Captain Lama May 7 at 13:38

## 1 Answer

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 $$s=\lambda$$).

• We can see how Hamilton (not yet knowing matrices, such as the Cayley-Hamilton theorem) took such a long time to do this. – GEdgar May 7 at 13:48
• To be honest I can't even see how Hamilton came up with these sort of questions in the first place. – Yorch May 7 at 13:54