Properties of cross product in $\mathbb{R^3}$

Question

• Does there exist $\mathbb{1}\in\mathbb{R^3}$ such that $\mathbb{1}\times X=X$ and $X\times\mathbb{1}=X$? ($\mathbb{1}$ as vector in $\mathbb{R^3}$)

My answer is no: because let $X=(1,2,3)$. Then we get,

$\mathbb{1}\times X=(1,1,1)\times (1,2,3)=(1,-2,1).$

Hence, we don't have $\mathbb{1}\times X=X$.

Can you check my answer?

Answer 1

The existence of $1\in \Bbb R^3$ depends on $X$.

For instance if $X=(0,0,0)$ then any $1\in \Bbb R^3$ satisfies your condition.

If $X\not=(0,0,0)$ then such a $1\in \Bbb R^3$ cannot exist.

Answer 2

Yes the answer is no but showing $(1,1,1)\times (1,2,3) \neq (1,2,3)$ is not sufficient since there may be other vector than $(1,1,1)$ as $1$ in $\mathbb{R}^3$.

Consider instead $1 = (x,y,z)$ if any, we want $$(x,y,z)\times (1,2,3) = (1,2,3)$$ or $$(3y - 2z,z - 3x, 2x - 3) = (1,2,3)$$ you can show that this equation doesn't hae solution for $x,y$ and $z$.

Answer 3

For any vectors $X$, $Y$, $X\times Y=-Y\times X$. If there were a vector such that $\mathbb 1\times X=X\times \mathbb 1=X$ for all $X$, then we would also have $X=-X$ for all $X$.

Answer 4

$\mathbb{1}$ is not always $(1,1,1)$, hence you cannot do this. Rather you must consider a arbitrary case.

Let $\mathbb{1}=(a,b,c)\in\mathbb{R}^3$. Let $X=(d,e,f)$

Suppose $\mathbb{1}\times X=X$, which means that $(bf-ec,-cd+fa,ea-db)=(d,e,f)$

Suppose $X\times\mathbb{1}=X$, which means that $(ec-bf, -af+cd, bd-ae)=(d,e,f)$

Adding these two equations would give us $(d,e,f)=X=0$ vector.

So therefore, your answer is that, if $X$ is the 0 vector, any $\mathbb{1}$ works, otherwise no.