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I want to prove the following identity: For $\mathbf a,\mathbf b\in \mathbb R^3, M\in M_3(\mathbb R)$, we have $$\mathbf b\times(\mathbf aM)-\mathbf a\times(\mathbf bM)=(\mathbf a\times\mathbf b)M-(\mathrm{tr} M )(\mathbf a\times\mathbf b)$$

One direct way to prove it is to write out the components of $\mathbf a,\mathbf b$ and $M$, then compare the result of $LHS$ and $RHS$ by brute force, but this way seems not so beautiful. Is there any nice (elegant) way to prove this identity? I guess probably there's a geometric interpretation of this identity.

FFjet
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    What is $M$? What is $a\times b$? If it is the cross product, then what is $M(a\times b)$? A vector? But then it cannot have a trace? – Dietrich Burde Jan 26 '21 at 14:27
  • There is still an issue with $(\mathbf a\times\mathbf b)M-(\mathrm{tr} M)(\mathbf a\times\mathbf b)$ as the first term is a matrix but the second one is a scalar. – Bill O'Haran Jan 26 '21 at 14:32
  • I have just edited the question, now it is clear. – FFjet Jan 26 '21 at 14:34
  • It might be useful to note that we can rewrite the left hand side as $$ (\mathbf{b-a} )\times (\mathbf aM) + \mathbf a \times (\mathbf a M - \mathbf b M). $$ – Ben Grossmann Jan 26 '21 at 14:46
  • Also, from this post, we have $$ (\mathbf a \times \mathbf b)M = \det(M)(\mathbf a M^{-T}) \times (\mathbf b M^{-T}) $$ – Ben Grossmann Jan 26 '21 at 14:54

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The proposed identity may be written as $$(\epsilon_{ijk}\delta_{ml} -\epsilon_{imk}\delta_{lj} -\epsilon_{lmj}\delta_{ki} +\epsilon_{imj}\delta_{lk})a_mb_jM_{lk}=0$$ where the expression in brackets is almost one of the identities listed in equations (5)-(9) in "The Isotropic Invariants of Fifth-rank Cartesian Tensors", Int. J. Quantum Chem. vol V 381-386 (1971).

Is the author of the question certain the first term on the right hand side of their identity is not $M \cdot ({\bf a}\times{\bf b})$, rather than $({\bf a}\times{\bf b})\cdot M$? (Assuming $M$ is not symmetric.) If it were, then the third term in the brackets would be $\epsilon_{kmj}\delta_{li}$, which yields a more symmetric result, and the expression in the brackets does vanish!

JMFG
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