Matlab has a function called planerot which takes a two-component column vector $x$ as input and returns a $2 \times 2$ matrix $G$ and a two component column vector $y$ such that $y = G.x$ and $y_2 = 0$. The documentation is available here.

I am trying to implement it using the ROTG routine of the BLAS package in Mathematica as follows

planerot[Xcol_] := Module[{a, b, a0, b0, c, s, y, givens},
  {a, b} = {a0, b0} = Xcol;
  ROTG[a, b, c, s];
  givens = {{c, s}, {-Conjugate[s], c}};
  y = givens . Transpose[{a0, b0}];
  {givens, y}]

I have tested this using the example given here and it works BUT if I pass a complex vector as the argument, Mathematica and Matlab gives different answers.

First, using $x = (3 ,4)^T$, we can see that Mathematica gives,

In[ ]:= {G, y} = planerot[{3, 4}] // N

Out[ ]= {{{0.6, 0.8}, {-0.8, 0.6}}, {5., 0.}}

which matches with the values given by MATLAB.

But using an example from Mathematica's documentation, our tiny method will give the following output for $x = (-2+3i,-2i)^T$,

In[808]:= {Gt, y} = planerot[{-2. + 3. I, -2. I}] // Chop 

Out[808]= {{{0.874475, -0.403604 - 0.269069 I}, {0.403604 - 
    0.269069 I, 0.874475}}, {-2.28709 + 3.43063 I, 0}}

whereas, MATLAB gives us

>> [G,y]= planerot([-2+3i,-2i]')

G =

  -0.4851 + 0.7276i   0.0000 - 0.4851i
   0.0000 - 0.4851i  -0.4851 - 0.7276i

y =


In both cases, $y_2 = 0$ but $y_1$ and the Givens matrix are different. How can I resolve this discrepancy? Or is there an open-source implementation of planerot that I can look at?

I would really appreciate your help.

  • 2
    $\begingroup$ Funny, at the moment Octave's planerot gives the same output as that of your Mathematica implementation: octave-online.net/… $\endgroup$
    – xzczd
    Feb 22, 2023 at 8:55
  • $\begingroup$ @xzczd I think the differences in the result arise from differences in definition. Although, thanks to the two answers here I can see how the differences arise, I am frankly a bit perplexed as to how to proceed. Which algorithm is "correct"? $\endgroup$
    – noir1993
    Feb 22, 2023 at 12:18
  • 2
    $\begingroup$ Hey, those solutions only differ by a multiplication with a complex rotation, i.e., with d = DiagonalMatrix[{Exp[I Arg[-2.28709 + 3.43063 I]], -I Exp[-I Arg[-0.403604-0.269069I]]}]. If you compute {{0.874475, -0.403604 - 0.269069 I}, {0.403604 - 0.269069 I, 0.874475}}.d and Conjugate[d].{-2.28709 + 3.43063 I, 0}, you'll get back MATLAB's solution. Now, which complex rotation is "correct" is a matter of taste. The same normalisation choice applies to eigenvalue decomposition, SVD, etc., where the solutions of Mathematica and MATLAB might vary. $\endgroup$ Feb 22, 2023 at 16:22

2 Answers 2


I don't know how this does make sense ($G$ is not orthogonal for the example complex input and $G.x \neq y$ !), but it seems to reproduce the results:

planerot[x : {x1_, x2_}] :=
 With[{norm = Norm[x]},
  {{{x1, x2},
    Conjugate[{-x2, x1}]}/norm,
   {norm, 0}}]

Frankly the fact that this function reproduces two results in the question doesn't cast much confidence in me for this being actually correct solution.

Numerical stability is a question I won't touch with a ten-foot pole, maybe you should check out the Wikipedia entry for Givens rotation.

  • 2
    $\begingroup$ Your function produces a matrix $G$ whose $G_{44}$ element is off by a sign. MATLAB gives -0.4851 - 0.7276i and your MMA function gives -0.485071 + 0.727607 i. I suppose taking a complex conjugate will fix the problem. $\endgroup$
    – noir1993
    Feb 22, 2023 at 12:30
  • 1
    $\begingroup$ @noir1993 Indeed! I have no idea why conjugates make this "work", but after that fix the matrix is at least unitary. I'm just wondering how results in general compare with the Matlab implementation... $\endgroup$
    – kirma
    Feb 22, 2023 at 12:47
  • 2
    $\begingroup$ @noir1993 and kirma, I just noticed we've all missed a feature of MATLAB core language, see the updated answer of mine. $\endgroup$
    – xzczd
    Feb 22, 2023 at 13:02
  • $\begingroup$ @xzczd Thank you! Great catch. Now my code gives me the correct $y$ but the Givens matrix is the complex conjugate of the one given by MATLAB. $\endgroup$
    – noir1993
    Feb 22, 2023 at 13:25

Definitely not the most efficient, but the most straightforward solution in my mind:

planerot = x |-> With[{y = {Norm@x, 0}}, {RotationMatrix@{x, y}, y}];

planerot@{3, 4.}
(* {{{0.6, 0.8}, {-0.8, 0.6}}, {5., 0}} *)

planerot@Conjugate@{-2. + 3 I, -2 I} // Chop
(* {{{-0.485071 + 0.727607 I,  0.       - 0.485071 I}, 
     { 0.       - 0.485071 I, -0.485071 - 0.727607 I}}, 
    {4.12311, 0}} *)

Notice I've added Conjugate in 2nd example, because in MATLAB ' is the shorthand for ctranspose (complex conjugate transpose). BTW, shorthand for transpose is .'.


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