I get stuck in the vector calculation from a textbook exercise(A Brief on Tensor Analysis) and hope someone can help me.
The problem is:
show that:
(1)$(\boldsymbol{uv})^T=\boldsymbol{vu}$
(2)for any vectors $\boldsymbol{a,b,c}$,$(\boldsymbol{a} \times \boldsymbol{b})\times \boldsymbol{c}=(\boldsymbol{ba}-\boldsymbol{ab})(\boldsymbol{c})$
How to prove the second statement?
What I have tried:
For the first problem, I set an arbitary vector $\boldsymbol{x}=(x_1, x_2, ..., x_n)$ and prove the equivalence with the matrix elements of the direct product of the 2 vectors $\boldsymbol{u,v}$.
For $\boldsymbol{uv}$, the ith row and jth column element is $(\boldsymbol{uv})_{[i,j]}=u_iv_j$
For $\boldsymbol{vu}$, the jth row and ith column element is $(\boldsymbol{vu})_{[j,i]}=v_ju_i=u_iv_j$
Therefore, for any vector $\boldsymbol{x}=(x_1, x_2, x_3, ..., x_n)$, $(\boldsymbol{uv})^T\boldsymbol{x}=(\boldsymbol{vu})\boldsymbol{x}$
Therefore, $(\boldsymbol{uv})^T=\boldsymbol{vu}$
Thank you!
After some investigation, I answer my question and hope it can help others.
First, $(\boldsymbol{a}\times\boldsymbol{b})\times\boldsymbol{c}=(\boldsymbol{a}\cdot\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{b}\cdot\boldsymbol{c})\boldsymbol{a}$ (as suggested by Suzu Hirose)
Then,$(\boldsymbol{b}\boldsymbol{a}-\boldsymbol{a}\boldsymbol{b})(\boldsymbol{c})=\boldsymbol{b}\boldsymbol{a}(\boldsymbol{c})-\boldsymbol{a}\boldsymbol{b}(\boldsymbol{c})$
Since $\boldsymbol{u}\boldsymbol{v}(\boldsymbol{w})=\boldsymbol{u}(\boldsymbol{v}\cdot\boldsymbol{w})$ (according to the properties of direct product)
We have $\boldsymbol{b}\boldsymbol{a}(\boldsymbol{c})-\boldsymbol{a}\boldsymbol{b}(\boldsymbol{c})=\boldsymbol{b}(\boldsymbol{a}\cdot\boldsymbol{c})-\boldsymbol{a}(\boldsymbol{b}\cdot\boldsymbol{c})$
Therefore $(\boldsymbol{a}\times\boldsymbol{b})\times\boldsymbol{c}=(\boldsymbol{b}\boldsymbol{a}-\boldsymbol{a}\boldsymbol{b})(\boldsymbol{c})$