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Is it true that for any two vectors $\mathbf{A}=A_1\mathbf{\hat{e}}_1+A_2\mathbf{\hat{e}}_2+A_3\mathbf{\hat{e}}_3=A'_1\mathbf{\hat{i}}+A'_2\mathbf{\hat{j}}_2+A'_3\mathbf{\hat{k}}$, $\mathbf{B}=B_1\mathbf{\hat{e}}_1+B_2\mathbf{\hat{e}}_2+B_3\mathbf{\hat{e}}_3=B'_1\mathbf{\hat{i}}_1+B'_2\mathbf{\hat{j}}+B'_3\mathbf{\hat{k}}$, where $\mathbf{\hat{e}}_1,\mathbf{\hat{e}}_2,\mathbf{\hat{e}}_3$ is an orthonormal basis of $\mathbf{R}^3$, $$\mathbf{A} \times \mathbf{B} = \det\begin{pmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\\ A'_1 & A'_2 & A'_3\\\ B'_1 & B'_2 & B'_3\end{pmatrix} = \det\begin{pmatrix}\mathbf{\hat{e}}_1 & \mathbf{\hat{e}}_2 & \mathbf{\hat{e}}_3\\\ A_1 & A_2 & A_3\\\ B_1 & B_2 & B_3\end{pmatrix}?$$ I think this is true but don't even know how to start the proof. I'd suspect this maybe has something to do with the fact that that the change of basis matrix between two orthonormal bases is orthonormal and that the cross product is invariant under orthogonal transformations but I'm not sure. I'd appreciate any tips / insights.

AJB
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  • Are you asking for a proof that the cross product is basis-independent? Or are you asking for a proof that the cross-product can be computed as a determinant in any orthonormal basis you choose? – march Oct 01 '24 at 18:43
  • Is this helpful? – J.G. Oct 01 '24 at 18:48
  • I vaguely suspect that this might not be true, because there's a sense in which basis vectors "rotate in the opposite direction" as the components of a vector, and so in going from one of the matrices in the post to the other, the transformation matrix is actually not orthogonal, and so the determinants might not be identical. So we should try to look for a counterexample – march Oct 01 '24 at 18:50
  • I will say what march said but more strongly: this is false since your orthogonal basis might have the wrong orientation. A simple counter example is $$e_1=i,\quad e_2=k,\quad e_3=j.$$ It will always be the same up to a sign, however. – Nicholas Todoroff Oct 01 '24 at 19:05
  • Ok, that makes sense. How would one justify that "it will always be the same up to a sign"? – AJB Oct 01 '24 at 19:22
  • You could prove the $\pm1$ factor is the orthogonal matrix's determinant, or you could restrict the proof to proper orthogonal matrices. – J.G. Oct 01 '24 at 19:34

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