How to Find the Rotation Matrix?

What is the possible solutions for an $3x3$ size orthogonal matrix with the known parameters of the first two columns which makes 6 parameters known?

Let's think of a rotation matrix which is an orthogonal matrix:

R=[r11 r12 r13;
   r21 r22 r23;
   r31 r32 r33]

The parameters of r11, r21, r31, r12, r22, r32 are known values. And it is desired to find the last (third) column with the parameters of r13,r23,r33.

I take the square root of each raw which is equivalent to 1 unit vector.\

r11^2+r12^2+r13^2=1
r21^2+r22^2+r23^2=1
r31^2+r32^2+r33^2=1

Therefore we can get to the equations of

r13= sqrt(1-r11^2-r12^2)
r23= sqrt(1-r21^2-r22^2)
r33= sqrt(1-r31^2-r32^2)

However, if there is one solution for the rotation matrix of R, how can I find that unique solution. Square root of any raw gives me two solutions.

Whichever (positive/negative square root of r13,r23,r33 I would take, it would be still an orthogonal matrix:

r1^2+r23^2+r33^2=1

So is there one unique solution or multiple solutions?

Answer 1

[David Goodmanson]

Hi ercan,

MODIFIED

First of all, for a rotation matrix the two known columns have to be an orthonormal pair, meaning that there are conditions on the initial six variables. If these variables are relabeled as two 3x1 column vectors c1,c2 then

norm(c1) = norm(c2) = 1;  dot(c1,c2) = 0

Let's assume that's true.

From your eqns, at first glance it might seem that there could be eight solutions, since you can have +-1 for each of r13,r23,r33. However, the rows have to be orthogonal, so e.g.

r11*r21 + r12*r22 + r13*r23 = 0

If you have a solution, then changing one of r13 or r23 will change the sign of the last term and will not provide a solution. You can change both signs or neither, but not just one. Similarly for the other pairs of rows. The net result is that there are two solutions. One solution is third column equaling the cross product of the first two columns (thus being perpendicular to both of them), and the other solution is simply that solution times an overall minus sign.

However, changing the overall sign in the third column also changes the sign of the determinant of the matrix. As Bruno pointed out, a rotation matrix must have determinant +1. So the only solution is

c3 = cross(c1,c2)

in which case

R = [c1 c2 cross(c1,c2)]

has determinant +1.

Note that for this to come out correctly there is a sign convention present (or one might say, hidden) in the cross function.

[ercan duzgun]

@David Goodmanson , thank you for your detailed answer.

[ercan duzgun]

@David Goodmanson, if I get the first solution of column 3, I can get the second solution by changing all signs of each matrix element in the 3rd column, right?

For example: if I have found

sol1=[ r_13_sol1;    r_23_sol1;    r_33_sol1]

Then the second solution would be

sol2=[ -1*r_13_sol1;    -1*r_23_sol1;    -1*r_33_sol1]

Is this right?

In your comment, you wrote:

"If you have a solution, then changing one of r13 or r23 will change the sign of the last term and will not provide a solution. You can change both signs or neither, but not just one. Similarly for the other pairs of rows. "

Therefore I was confused a bit. Because you say: changing one of r13 or r23, but I find that we should change all three elements of the column. Could you explain this ?

[ercan duzgun]

clear all;clc;
syms r11 r21 r31 r12 r22 r32 r13 r23 r33 real
R=[r11 r12 r13;
   r21 r22 r23;
   r31 r32 r33];
sol1=cross(R(1:3,1),R(1:3,2))
solution_R_matrix1=[R(1:3,1:2) sol1]

sol2=cross(R(1:3,1),R(1:3,2))*-1
solution_R_matrix2=[R(1:3,1:2) sol2]

@David Goodmanson , is this code correct for the two solutions?

[David Goodmanson]

Hi ercan,

What I meant by 'similarly for the other pairs of rows' is that what is true for rows 1 and 2 is also true for rows 1 and 3

r11*r31 + r12*r32 + r13*r33 = 0

so by the same argument you can change either neither or both signs of r13,r33, but not just one sign. Combining that with the result for r13,r23 means that all three signs have to change simlutaneously.

Answer 2

[Bruno Luong]

You equation is NOT enough to determine the third column, you forget about the orthogonal conditions and that det(R)=1 (rotation matrix is SO(3)).

So solutions is simply

R(:,3) = cross(R(:,1),R(:,2))

[ercan duzgun]

Dear @Bruno Luong thank you for your response.

One solution is this in your code:

R(:,3) = cross(R(:,1),R(:,2))
solution1_column=cross(R(:,1),R(:,2))

According to @David Goodmanson's reply, should I take the minus sign of all of the elements for another solution(second solution)?

solution2_column=-1*solution1_column

And is it sure that there wouldn't be any other solution, but only two solutions?

[Bruno Luong]

One solution (since det R=1). SO(3) has 6 DOFs that can be uniquely derived from 2 columns, or 2 rows of R.

If you inverse the sign, you'll get a "mirror" like transformation. This matrix is NOT a rotation matrix

[1  0  0;
 0  1  0;
 0  0 -1];

[David Goodmanson]

Hi ercan,

What Bruno pointed out is correct, and I have modified the answer accordingly.

[ercan duzgun]

Thank you @David Goodmanson , thank you @Bruno Luong .