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Solve a system of linear equations?

Posted 8 years ago

Hi!

I have a pretty easy problem where I have 6 equations and 6 unknown of the form m*x = 0, where m is a 6x6 matrix and x is a vector of length 6 with the 6 unknown variables. I'm looking for a nontrivial solution to this problem so I use the Nullspace command.

My code is the following

m = {{E^(-I*x*b/2), E^(I*x*b/2), -E^(-I*y*b/2), -E^(I*y*b/2), 0, 0},
 {I*x*E^(-I*x*b/2), -I*x*E^(I*x*b/2), -I*y*E^(-I*y*b/2), 
  I*y*E^(I*y*b/2), 0, 0},
 {0, 0, -E^(I*y*b/2), -E^(-I*y*b/2), E^(I*x*b/2), E^(-I*x*b/2)},
 {0, 0, -I*y*E^(I*y*b/2), I*y*E^(-I*y*b/2), 
  I*x*E^(I*x*b/2), -I*x*E^(-I*x*b/2)},
 {E^(-I*x*(a + b/2)), E^(I*x*(a + b/2)), 0, 0, 0, 0},
 {0, 0, 0, 0, E^(I*x*(a + b/2)), E^(-I*x*(a + b/2))}}
ns = NullSpace[m]

This gives me the output { }. What does this result mean? All the coefficients x,y,a,b are real and greater than zero. Is that something that I should specify in the code? I wonder what is wrong with my code?

I've attached my code to this post.

Thanks in advance!

Attachments:
POSTED BY: Pontus Vikstål
7 Replies

NullSpace gives you a basis for the null space. When the null space is is zero-dimensional, its basis is the empty set {}.

You get a nonempty NullSpace for special values of the parameters:

sols = Solve[Det[m] == 0]
MatrixForm[m] /. sols
Map[NullSpace, m /. sols]
POSTED BY: Gianluca Gorni
Attachments:
POSTED BY: Russell Michael
POSTED BY: Gianluca Gorni

Hi, I have the same problem and I would like to know how to solve such a system of linear equations

POSTED BY: Russell Michael

No nulls is good nulls?

POSTED BY: Daniel Lichtblau

For m *x == 0 you must have Det[ m ] == 0, but

In[17]:= Det[m] // FullSimplify

Out[17]= E^(-I (2 a x + b y)) (-(x + E^(2 I a x) (x - y) + y)^2 +    E^(2 I b y) (x - y + E^(2 I a x) (x + y))^2)
POSTED BY: Hans Dolhaine
Posted 8 years ago

Perhaps

MatrixForm[RowReduce[m]]

and

NullSpace[IdentityMatrix[6]]

might give a hint

POSTED BY: Bill Simpson
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