Help over the proof of triple vector product identity

Question

For all vectors $\bf{x}$, $\bf{y}$ and $\bf{z}$,

$$\bf{x}\times(\bf{y}\times\bf{z})=(\bf{x}\cdot\bf{z})\bf{y}-(\bf{x}\cdot\bf{y})\bf{z}$$

The proof goes as follows:

We may suppose that $\bf{y}$ and $\bf{z}$ lie along different lines through the origin, for otherwise it is is trivially true. Then any $\bf{x}$ can be written as a linear combination of $\bf{y}$,$\bf{z}$ and $\bf{y}\times\bf{z}$. As both sides are linear in $\bf{x}$, it is only necessary to verify the identity when $\bf{x}$ is each of these three vectors.

$\bigg[$1. My first question, why? Why is it only necessary to verify these three possibilities?$\bigg]$

When $\bf{x}=\bf{y}\times\bf{z}$ all terms are zero, so it is true. The two cases $\bf{x}=\bf{y}$ and $\bf{x}=\bf{z}$ are the same (apart from a factor −1 on each side of the equation), so we have reduced our task to giving a proof of the identity

$$\bf{y}\times(\bf{y}\times\bf{z})=(\bf{y}\cdot\bf{z})\bf{y}-(\bf{y}\cdot\bf{y})\bf{z}$$

for all $\bf{y}$ and $\bf{z}$.

Now the latter is true when $\bf{y}$ and $\bf{z}$ lie along the same line through the origin. Thus, as both sides of the latter identity are linear in $\bf{z}$, it only remains to prove this when $\bf{z}$ is orthogonal to $\bf{y}$.

$\bigg[$2. Now few questions at this stage: how does the author determines that both sides are linear in $\bf{z}$ and why does it only remain to prove it when $\bf{z}$ is orthogonal to $\bf{y}$?$\bigg]$

Moreover, it is clearly sufficient to prove this when $\bf{y}$ and $\bf{z}$ are unit vectors. Now $\bf{y}\times(\bf{y}\times\bf{z})$ is of length one and also a scalar multiple of $\bf{z}$. Thus it is $\pm\bf{z}$, so if we let $\mu=\bf{z}\cdot[\bf{y}\times(\bf{y}\times\bf{z})]$, then $\mu=\pm1$. However, $\mu$ is clearly a continuous function of the coefficients of $\bf{y}$ and $\bf{z}$, so it must be independent of $\bf{y}$ and $\bf{z}$.

$\bigg[$3. What do he mean by clearly continuous function independent of $\bf{y}$ and $\bf{z}$?$\bigg]$

As $\mu = −1$ when $\bf{y}=\bf{i}$ and $\bf{z}=\bf{j}$ , we see that $\bf{y}\times(\bf{y}\times\bf{z})=-\bf{z}$ and the proof is complete.

Thanks in advance!

Answer 1

1.) In $\mathbf{R}^3$ the vector system $\bf{y},\bf{z},\bf{y}\times\bf{z}$ is a basis (supposing that $\bf{y}$ and $\bf{z}$ lie along different lines through the origin). So if you want to check the equality of two linear maps it is enough to check for basis vectors.

2.) Putting $\bf{z}:=\bf{z}_1+\alpha\bf{z}_2$ and using the linearity of cross product (vector product) and inner product you obtain the linearity of the mentioned expressions. When $\bf{y}$ and $\bf{z}$ do not lie along the same line through the origin, then you can decompose (uniquely) $\bf{z}=\bf{z}_{||}+\bf{z}_{\perp}$, where $\bf{z}_{||}$ is parallel to $\bf{y}$ and $\bf{z}_{\perp}$ is perpendicular (orthogonal) to $\bf{y}$.

3.) Sorry :-) I don't understand this continuity argument. Hopefully someone else can help you to make clear this last step.

Answer 2

I'll help out with #3, since vesszabo gave satisfactory answers to the first two.

Since addition and multiplication operations in $\Bbb R^3$ are continuous operations, then any manner of finite compositions of these operations creates a continuous map. So, what we have here is a continuous map $U \to \{-1,1\}$ where $U$ is the unit sphere from the unit sphere $U\subset\Bbb R^3$.

But, the author notes, you have an example of something that maps to $-1$. Therefore everything in $U$ maps to $-1$: otherwise you have discontinuity. $U$ is a connected set, and a continuous image of $U$ can't get split into two pieces going into both $-1$ and $1$.