I want to solve these equations using MATLAB and I am sure there is a non zero solution. The equations are:
0.7071*x + 0.7071*z = x
-0.5*x + 0.7071*y + 0.5*z = y
-0.5*x - 0.7071*y + 0.5*z = z
I wrote in MATLAB:
[x,y,z]=solve('0.7071 * x+0.7071 * z=x','-0.5 * x+0.7071 * y+0.5 * z=y','-0.5 * x-0.7071 * y+0.5 * z=z');
But the result is x = y = z = 0. As I said I am sure that there is a solution. Can any one help?
You're looking for a non-trivial solution v to A*v=v with v=[x;y;z] and...
A =
0.70710678118655 0 0.70710678118655
-0.50000000000000 0.70710678118655 0.50000000000000
-0.50000000000000 -0.70710678118655 0.50000000000000
You can transform this into (A-I)v=0 where I is the 3x3 identity matrix. What you have to do to find a nontrivial solution is checking the null space of A-I:
>> null(A-eye(3))
ans =
0.67859834454585
-0.67859834454585
0.28108463771482
So, you have a onedimensional nullspace. Otherwise you'd see more than one column. Every linear combination of the columns is a point in this null space that A-I maps to the null vector. So, every multiple of this vector is a solution to your problem.
Actually, your matrix A is a rotation matrix of the first kind because det(A)=1 and A'*A=identity. So it has an eigenvalue of 1 with the rotation axis as corresponding eigenvector. The vector I computed above is the normalized rotation axis.
Note: For this I replaced your 0.7071 with sqrt(0.5). If rounding errors are a concern but you know in advance that there has to be a nontrivial solution the best bet is to do a singular value decomposition of A-I and pick the right most right singular vector:
>> [u,s,v] = svd(A-eye(3));
>> v(:,end)
ans =
0.67859834454585
-0.67859834454585
0.28108463771482
This way you can calculate a vector v that minimizes |A*v-v| under the constraint that |v|=1 where |.| is the Euclidean norm.
I don't think you need to use the solve function as your equations is a system of linear equations.
Recast as a matrix equation:
Ax = B
In your case:
| -0.2929 0.0 0.7071 | | x | | 0 |
| -0.5 -0.2929 0.5 | | y | = | 0 |
| -0.5 -0.7071 -0.5 | | z | | 0 |
Use the built-in functionally of MATLAB to solve it. See e.g. MATLAB: Solution of Linear Systems of Equations.
The core of MATLAB is to solve this kind of equation.
Using FreeMat (an open-source MATLAB-like environment with a GPL license; direct download URL for Windows installer):
A = [ -0.2929 0.0 0.7071; -0.5 -0.2929 0.5; -0.5 -0.7071 -0.5 ]
B = [0.0; 0.0; 0.0]
A\B
ans =
0
0
0
So the solution is: x = 0, y = 0, z = 0
The solution can also be derived by hand. Starting from the last two equations:
-0.5*x + 0.7071*y + 0.5*z = y
-0.5*x - 0.7071*y + 0.5*z = z
0.2929*y = -0.5*x + 0.5*z
0.7071*y = -0.5*x + 0.5*z
0.2929*y = 0.7071*y
Thus y = 0.0 and:
0.7071*y = -0.5*x + 0.5*z
0 = -0.5*x + 0.5*z
0 = -0.5*x + 0.5*z
0.5*x = 0.5*z
x = z
Inserting in the first equation:
0.7071*x + 0.7071*z = x
0.7071*x + 0.7071*x = x
1.4142*x = x
Thus x = 0.0. And as x = z, then z = 0.0.
x = 0, y = 0, z = 0 is the correct solution. This problem can be easily worked by hand.
A = [ 0.7071 0 0.7071 ;
-0.5 0.7071 0.5 ;
-0.5 -0.7071 0.5 ];
B = [1 ; 1 ; 1];
AA = A-diag(B)
-0.2929 0 0.7071
-0.5 -0.2929 0.5
-0.5 -0.7071 -0.5
BB = B-B
0
0
0
AA\BB
0
0
0
Friends use MATLAB command rref(A) for Ax=B.... It will give a upper triangular Matrix. Then u can calculate non-trivial solutions easily... But keep in mind that if rref(A)=I (as in your case) then non-trivial solutions do not exist. BEST OF LUCK
