Split Jacobian result into Matrix factors

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Simon Detmer 2020-5-25

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https://ww2.mathworks.cn/matlabcentral/answers/532773-split-jacobian-result-into-matrix-factors

评论： David Goodmanson 2020-5-27

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Hello,

I'm using the symbolic toolbos to calculate the following Jacobian:

jacobian(A4,phi_)

where

A4 = - TLB*R*TLB2*w

with

The result of jacobian(A4,phi_) calculated by Matlab is

[wz*(r2 + psi*r5 - r8*theta) - wy*(r3 + psi*r6 - r9*theta),   wy*(r8 + phi*r9 - psi*r7) - wz*(r1 - r9 + phi*r8 + psi*r4 - 2*r7*theta) + wx*(r3 + r7 + psi*r6 + psi*r8 - 2*r9*theta), - wz*(r6 - phi*r5 + r4*theta) - wy*(r5 - r1 + phi*r6 - 2*psi*r4 + r7*theta) - wx*(r2 + r4 + 2*psi*r5 - r6*theta - r8*theta)]
[   - wx*(r7 + psi*r8 - r9*theta) - wz*(r9 - r5 - 2*phi*r8 + psi*r2 + r7*theta) - wy*(r6 + r8 + 2*phi*r9 - psi*r3 - psi*r7),wx*(r6 + phi*r9 - psi*r3) - wz*(r4 + phi*r7 - psi*r1),       wz*(r3 - phi*r2 + r1*theta) - wx*(r5 - r1 + phi*r8 - 2*psi*r2 + r3*theta) + wy*(r2 + r4 + phi*r3 + phi*r7 - 2*psi*r1)]
[ wx*(r4 + psi*r5 - r6*theta) - wy*(r9 - r5 - 2*phi*r6 + psi*r4 + r3*theta) + wz*(r6 + r8 - 2*phi*r5 + r2*theta + r4*theta), - wy*(r2 + phi*r3 - psi*r1) - wx*(r1 - r9 + phi*r6 + psi*r2 - 2*r3*theta) - wz*(r3 + r7 - phi*r2 - phi*r4 + 2*r1*theta), wy*(r7 - phi*r4 + r1*theta) - wx*(r8 - phi*r5 + r2*theta)]

which as you can see isn't very nice looking.

My question is if I can somehow turn this result into a product of my original matrices (if even mathematically possible),

for example ans = TLB*TLB2*R^T*inv(TLB*TLB2) or something like that.

Hope someone can help me here, thanks.

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Simon Detmer 2020-5-26

So no possibility to simplify the result while keeping the symbolic variables/matrices? shame...

Thank you for your answer though.

David Goodmanson 2020-5-27

With fifteen independent variables sprinkled around by matrix multiplication, you can't expect the results to be sleek. And if you do have a better looking alternative expression such as the one you suggest (assuming it were true) then of course the end result has to be exactly the same as the one you have. So a better starting point doesn't help.

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