Finding Null Space of a Matrix with Trigonometric Equations

[indigojoker]

I need to find the null space of:

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

so:
\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

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

for a while now and I'm not sure how to get the x and y values

 

[Hurkyl]

It's just a linear system of equations, isn't it? How do you normally solve systems of linear equations?

 

[indigojoker]

by substitution, but the thing is I keep getting y=0 and x=0

 

[genneth]

Does the obvious x = -sin(\beta) e^{-ia}, y = cos(\beta)-1 not work?

 

[indigojoker]

sure, but i would like to know how to solve for that.

 

[Pseudo Statistic]

I need to find the null space of:

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

so:
\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

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

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 x, you get:
x=-\frac{sin(\beta)e^{-i \alpha}}{cos(\beta)-1}y
And hence, you have \binom{-\frac{sin(\beta)e^{-i \alpha}}{cos(\beta)-1}y}{y} 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.

 

[genneth]

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.

 

[indigojoker]

thanks for the advice!

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

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

I got eigenvalues of +1 and -1

if we took the -1 case, we have:

\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

my null space would be:

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

However, this does not satisfy the eigenvalue equation: Ax=x\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}}

Not sure what this means

 

[genneth]

It looks correct to me... The relevant equation is not Ax=x, but Ax = ax, where a is the eigenvalue.