0

Before you read my question please consider that i HAVE to do this exercise as i did below.

So i am showing there exists no division algebra on $\mathbb R^3$.

To show that, i have to show that for $*: \mathbb R^3 \times \mathbb R^3 \to \mathbb R^3$ it exists $0 \ne x,y \in \mathbb R^3$ with $x*y = 0$ $(1)$

I first showed that there exists a linear matrix $A(x) \in \mathbb R^{3 \times 3}$ with $A: x \to A(x)$.

Then i did show that $det(A(x)) = 0$ for some $x \in \mathbb R^3 \setminus \{0\}$

At last i defined $L_x:\mathbb R^3 \to \mathbb R^3, y \to x*y$ and i showed it is linear and it satisfies $L_x(y) = A(x).y$.

Now to my problem:

I have to use what i did here and the fact that a Matrix $A$ is invertible if $Ay \ne 0$ for all $y \in \mathbb R^3 \setminus \{0\}$.

How and Why follows $(1)$ from what i did here.

I am unable to make a connection between what i did and what i shall show ( maybe because i am still newbie ).

El magnifico
  • 87
  • See this post. You $L_x$ is there also used. – Dietrich Burde Jul 06 '21 at 16:59
  • @DietrichBurde Thank you, but i saw it before. The Problem is that i am not allowed to use much more than i did show above ... – El magnifico Jul 06 '21 at 17:02
  • From linear algebra you should recall that $\det A=0$ implies non-trivial kernel. So there exists $y\neq0$ such that $L_x(y)=0$. Done. Mind you, the order in which you list the facts you have done feels a bit off. You cannot prove those things in that order. You need to begin from the definition of $L_x$ for a carefully chosen $x$ (that should not be a scalar multiple of the identity). – Jyrki Lahtonen Jul 11 '21 at 02:51

0 Answers0