Meaning of the symbol '' \ ''

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Waqas Syed 2015-7-28

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编辑： Bruno Luong 2021-10-1

采纳的回答： Lukas Bystricky

Hi,

I am trying to understand a pre written matlab code. There is a line: dx = -(J \ F);

What does the forward slash mean? Does it mean F divided by J or something?

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采纳的回答

Lukas Bystricky 2015-7-28

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编辑：Lukas Bystricky 2015-7-28

That's actually a backward slash. J\F is equivalent to solving the system J dx = F for dx, where J is a matrix (probably a Jacobian) and F and dx are vectors.

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Walter Roberson 2021-5-31

在 MATLAB Online 中打开

@Jeong Cheol Park

Could you give an example? Testing seems to indicate that the two are equivalent to within round-off error.

format long g
A = complex(randn(5,5), randn(5,5))
A = 
       0.000992623799124492 +    0.0560829640495359i         -0.905381318777621 -     0.398393560374225i          0.700377093990676 -     0.862124510784948i          0.404304943943166 -     0.958126231680329i         -0.890974479421001 -     0.621663836656907i
           -1.9302742743487 +     0.406700314684335i         -0.304220505361875 -     0.811612857394268i          0.651108834598317 -     0.230628781469304i        -0.0444221267375187 -      1.30394069512746i          -1.70637813489689 +      1.36627116598296i
         0.0164551669159734 +     0.528580646281319i          0.706575845098412 -      1.50266720293766i         -0.769410875922517 +     0.141780402125759i         -0.682477282104919 +      1.32055793476325i           -2.1534998275902 +     0.799074214788328i
          -1.10561081495984 +      1.19076502024226i          0.599408664135243 -     0.113702650020593i          -1.20756535401639 +      1.81944185552293i         -0.459403598927519 +     0.316798408043163i          0.934795804814805 +     0.738804986955679i
         -0.480841862945596 +     0.217646340463854i         0.0147301249197675 -     0.246444159536535i           1.59613276808549 +     0.699415876558451i          -1.57319944252273 +     0.112913838851912i          0.788541640989461 +      0.48711745728954i
B = complex(randn(5,1),randn(5,1))
B = 
          -0.20671001769678 -     0.721019480051865i
         -0.613706996033907 -    0.0545996685571732i
         -0.669498867822517 +      1.22665047948491i
           1.06282581209151 -      0.43106669371497i
          0.894215103837536 +      1.41339741033932i
C = A\B
C = 
         -0.981249160199436 +     0.466090069260794i
          -1.21058068247188 +     0.473613281424541i
         -0.304857696301016 -     0.125681359475334i
         -0.307334851145085 -     0.808043204705325i
          0.638459642251516 +     0.522575122891811i
D = inv(A)*B
D = 
         -0.981249160199436 +     0.466090069260794i
          -1.21058068247188 +     0.473613281424541i
         -0.304857696301016 -     0.125681359475334i
         -0.307334851145085 -     0.808043204705325i
          0.638459642251515 +     0.522575122891811i
E = C-D
E = 
      -1.11022302462516e-16 +                     0i
       2.22044604925031e-16 +                     0i
       5.55111512312578e-17 -  1.11022302462516e-16i
                          0 +  3.33066907387547e-16i
       3.33066907387547e-16 +                     0i
F = sum(E(:).^2)
F = 
       5.23852944873328e-32 -  1.23259516440783e-32i

Bruno Luong 2021-10-1

在 MATLAB Online 中打开

@Walter Roberson: "So J\F vs F/J would only be the same if F and J are each real and symmetric."

Not true

J=rand(3); J=J+J',
J = 3×3
    0.0670    0.8628    0.2369
    0.8628    1.6893    1.1714
    0.2369    1.1714    1.6260
F=rand(3); F=F+F',
F = 3×3
    0.8988    0.3888    0.1346
    0.3888    0.2691    0.9150
    0.1346    0.9150    1.4615
J\F
ans = 3×3
   -1.1987   -1.0761   -0.0648
    1.3265    0.4196   -0.1101
   -0.6982    0.4172    0.9877
F/J
ans = 3×3
   -1.1987    1.3265   -0.6982
   -1.0761    0.4196    0.4172
   -0.0648   -0.1101    0.9877

Bruno Luong 2021-10-1

编辑：Bruno Luong 2021-10-1

@Kriti Salwan "What happens if we are applying backlash to a matrix which is not square ?"

x = A\b;

For A with rank(A) < size(A,2) see https://blogs.mathworks.com/cleve/2021/04/28/solving-commodious-linear-systems/

Otherwise it's a least square solution

x = argmin(norm(A*x-b))

or equivalently

x = pinv(A)*b

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