Assume input as follows

in[1]=w = {a, b, c}; v = {d, e, g};
jMatrix = {{J1, 0, 0}, {0, J2, 0}, {0, 0, J3}};
ans= 1/2*jMatrix.Cross[v, w] + 1/2*(Cross[v, jMatrix.w] + Cross[w,jMatrix.v]) + jMatrix.Cross[w, v];

we have output[1]

out[1]={-(1/2) (c e - b g) (j1 + j2 + j3), 1/2 (c d - a g) (j1 + j2 + j3), -(1/2) (b d - a e) (j1 + j2 + j3)}

Can i simplify the out[1] to vector crossproduct or dotproduct form(in other words, more concise form.)?

ans1 // Simplify seems don't work.

Thanks @Carl Woll, TensorReduce works well, but it failed when i set jMatrix as $I_3$.

ans1 = 1/2*jMatrix.Cross[v, w] + 
   1/2*(Cross[v, jMatrix.w] + Cross[w, jMatrix.v]) + 
   jMatrix.Cross[w, v];
 Assumptions -> (v | w) \[Element] Vectors[3] && 
   jMatrix \[Element] IdentityMatrix[3]]//TeXForm

the result output not concise.

  • $\begingroup$ If mma can specify the form, and then to find the coefficient, if expression don't match the hypothetical form, only add some residual part.... $\endgroup$ – Ben Dec 19 '17 at 19:30

If you avoid using explicit vectors and matrices, you could use TensorReduce. Here is your expression:

ans //TeXForm

$\frac{1}{2} (v\times (\operatorname{jMatrix}.w)+w\times (\operatorname{jMatrix}.v))+\frac{\operatorname{jMatrix}.v\times w}{2}+\operatorname{jMatrix}.w\times v$

And here is the result of TensorReduce:

    Assumptions -> (v|w) ∈ Vectors[3] && jMatrix ∈ Matrices[{3,3}]
] //TeXForm

$\frac{1}{2} v\times (\operatorname{jMatrix}.w)+\frac{1}{2} w\times (\operatorname{jMatrix}.v)-\frac{\operatorname{jMatrix}.v\times w}{2}$

Slightly simpler than your original expression.


The OP asked about setting jMatrix to IdentityMatrix[3]. For this you could use my TensorSimplify paclet. Install the paclet with:

    "Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"

Once installed, you can load the package with:


Then, the following does the simplification you want:

$Assumptions = (v|w) ∈ Vectors[3];

TensorSimplify[ans1 /. jMatrix -> Inactive[IdentityMatrix][3]] //TeXForm

$\frac{w\times v}{2}$

  • $\begingroup$ Let me try, if i find some technology, i will update this question, thank you for your patience.@Carl Woll $\endgroup$ – Ben Dec 19 '17 at 19:35
  • $\begingroup$ Amazing! thank you again@Carl Woll $\endgroup$ – Ben Dec 19 '17 at 19:42
  • $\begingroup$ good job! @Carl Woll $\endgroup$ – Ben Dec 20 '17 at 1:48


ClearAll[w, v, jMatrix]
Inactivate[1/2  jMatrix . Cross[v, w] + 1/2 (Cross[v, jMatrix . w] + 
 Cross[w, jMatrix . v]) + jMatrix . Cross[w, v], Cross|Dot]

$\frac {1} {2} (v\times (\text {jMatrix}.w) + w \times (\text {jMatrix}.v)) + \frac {\text {jMatrix}.v\times w} {2} + \text {jMatrix}.w\times v$

  • $\begingroup$ it can works, thank you! @kglr $\endgroup$ – Ben Dec 20 '17 at 1:49
  • $\begingroup$ @Ben, my pleasure. $\endgroup$ – kglr Dec 20 '17 at 1:53

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