Finding Null Space of a Matrix with Trigonometric Equations

In summary, the two equations are not independent, so the solution, if one exists, would be a line. Such a line has to include the origin, so just eliminate/ignore one of the equations, and what you've got left should describe the line.
  • #1
indigojoker
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0
I need to find the null space of:

[tex]\dotx \left(\begin{array}{cc}cos(\beta)-1&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)-1\end{array}\right)[/tex]

so:
[tex]\dotx \left(\begin{array}{cc}cos(\beta)-1&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)-1\end{array}\right) \binom{x}{y} = 0[/tex]

I'm not sure how to go about doing this because I've been staring at:
[tex](cos(\beta)-1)x=-sin(\beta)e^{-i \alpha} y[/tex]
[tex]sin(\beta) e^{i \alpha} x = (-cos(\beta)-1) y[/tex]

for a while now and I'm not sure how to get the x and y values
 
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  • #2
It's just a linear system of equations, isn't it? How do you normally solve systems of linear equations?
 
  • #3
by substitution, but the thing is I keep getting y=0 and x=0
 
  • #4
Does the obvious [tex]x = -sin(\beta) e^{-ia}, y = cos(\beta)-1[/tex] not work?
 
  • #5
sure, but i would like to know how to solve for that.
 
  • #6
indigojoker said:
I need to find the null space of:

[tex]\dotx \left(\begin{array}{cc}cos(\beta)-1&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)-1\end{array}\right)[/tex]

so:
[tex]\dotx \left(\begin{array}{cc}cos(\beta)-1&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)-1\end{array}\right) \binom{x}{y} = 0[/tex]

I'm not sure how to go about doing this because I've been staring at:
[tex](cos(\beta)-1)x=-sin(\beta)e^{-i \alpha} y[/tex]
[tex]sin(\beta) e^{i \alpha} x = (-cos(\beta)-1) y[/tex]

for a while now and I'm not sure how to get the x and y values
Usually when you solve a problem like this-- i.e. a homogeneous linear system, you're going to end up with the trivial solution (x,y)=(0,0) and a surplus of other solutions with a free parameter.
What you usually do is solve one of these equations and use a variable as a parameter.
For example, if you solve the first equation you've been staring at for [tex]x[/tex], you get:
[tex]x=-\frac{sin(\beta)e^{-i \alpha}}{cos(\beta)-1}y[/tex]
And hence, you have [tex]\binom{-\frac{sin(\beta)e^{-i \alpha}}{cos(\beta)-1}y}{y}[/tex] as your null-space, y being a free parameter.
You could have just as well have done it for x, with that as the free parameter.
Ofcourse, you could always simplify this.
You can do similarly for the second equation you got and obtain a basis for your null space.
 
  • #7
indigojoker said:
sure, but i would like to know how to solve for that.

Notice that the two equations are not independent, so the solution, if one exists, would be a line. Such a line has to include the origin, so just eliminate/ignore one of the equations, and what you've got left should describe the line.
 
  • #8
thanks for the advice!

In the grand scheme of things, I was trying to find the eigenvectors of:

[tex]A=\dotx \left(\begin{array}{cc}cos(\beta)&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)\end{array}\right)[/tex]

I got eigenvalues of +1 and -1

if we took the -1 case, we have:

[tex]\dotx \left(\begin{array}{cc}cos(\beta)+1&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)+1\end{array}\right) \binom{x}{y} = 0[/tex]

my null space would be:

[tex]x=\binom{-sin(\beta)}{(cos(\beta)+1)e^{i \alpha}}[/tex]

However, this does not satisfy the eigenvalue equation: Ax=x[tex]\dotx \left(\begin{array}{cc}cos(\beta)-1&sin(\beta)e^{-i \alpha}\\sin(x)e^{i \alpha}&-cos(\beta)-1\end{array}\right) \binom{-sin(\beta)}{(cos(\beta)+1)e^{i \alpha}}=\binom{sin(\beta)}{(-cos(\beta)-1)e^{i \alpha}}[/tex]

Not sure what this means
 
  • #9
It looks correct to me... The relevant equation is not Ax=x, but Ax = ax, where a is the eigenvalue.
 

1. What is null space?

Null space, also known as kernel, is a vector space that contains all vectors that when multiplied by a particular matrix, result in the zero vector. In other words, it is the set of all solutions to a homogeneous equation Ax = 0.

2. How is null space related to linear transformations?

Null space is related to linear transformations as it represents the set of vectors that are mapped to the zero vector by the transformation. It is also known as the "vanishing set" of the transformation.

3. How do you find the null space of a matrix?

The null space of a matrix can be found by solving the homogeneous equation Ax = 0. This can be done by reducing the matrix to row echelon form and identifying the free variables in the resulting system of equations. The null space will then be spanned by the columns corresponding to the free variables.

4. What is the dimension of the null space?

The dimension of the null space is also known as the nullity of a matrix and is equal to the number of free variables in the reduced row echelon form of the matrix. It can also be calculated as the number of columns of the matrix minus the rank of the matrix.

5. Why is the null space important?

The null space is important because it helps us understand the behavior of a linear transformation. It provides information about the solutions to a system of linear equations and can be used to find a basis for the null space, which is useful in solving many engineering and scientific problems.

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