On page 13 of doCarmo's book on Differential geometry of curves and surfaces he stated that:

...we prove the relation:

$\displaystyle (u \wedge v)\cdot(x \wedge y) = \left | \begin{array}{cc} u\cdot x & v\cdot x \\ u \cdot y & v \cdot y \end{array} \right | $

where all $\displaystyle u,v,x,y$ are arbitrary vectors. This can easily be done by observing that both sides are linear in $\displaystyle u,v,x,y$ Thus it suffices to check that:

$\displaystyle (e_i \wedge e_j)\cdot (e_k \wedge e_l) = \left | \begin{array}{cc} e_i\cdot e_k & e_j\cdot e_k \\ e_i \cdot e_l & e_j \cdot e_l \end{array} \right | $

for all $\displaystyle i,j,k,l = 1, 2, 3$

My question: what does it mean when the author said that

"$\displaystyle \text{... can easily be done by observing that both sides are linear}$"? What did he mean by the word linear?

Also the author said that:

$\displaystyle \text{Thus it suffices to check that:}$

$\displaystyle (e_i \wedge e_j)\cdot (e_k \wedge e_l) = \left | \begin{array}{cc} e_i\cdot e_k & e_j\cdot e_k \\ e_i \cdot e_l & e_j \cdot e_l \end{array} \right |$

where he defined $\displaystyle e_1 \text{ as } (1,0,0)\,\, e_2 \text{ as } (0,1,0) \text{ and }\,\, e_3 \text{ as } (0,0,1)$?

Why is it sufficient to prove the above statement of the author for standard basis $\displaystyle (1,0,0), (0,1,0), (0,0,1)$ only?

How does this proof using standard basis automatically validates for non-standard basis? I know the author is right. But isn't it something like choosing a particular basis to prove our theorem which can be false for other non-standard bases?

I can't find the answer. Is it possible to kindly help me detect what I'm lacking in understanding this statement?