Eigen-things.

May 25, 2010

This is going to be a short post about how to find an eigenvalue and its corresponding eigenvectors.  Okay, for a matrix A and a vector x, we want to find a \lambda such that Av = v\lambda.  This means that when we multiply v by some matrix, we get a scalar multiple of that same matrix!  Pretty sweet.  How do we find such a thing?

Well, let’s think about this.  If Av = v\lambda then we have Av - v\lambda = 0, which means, when we factor out a v we get v(A - I\lambda)v = 0 with I is the identity matrix.  Since, by definition, we do not care when v = 0, we have that such a \lambda exists when we have A - I\lambda = 0.  This is slightly difficult to compute, of course, so we must be clever and take the determinant of both sides.  We get det(A - I\lambda) = 0.

Any \lambda such that this holds is called an eigenvalue, and the associated v is called the corresponding eigenvector.  This works for any square matrix, of any size.  The determinant is much more difficult to take, but there’s an easy way to manipulate this to make it nicer.  Either way, for the 2×2 form, it’s quite simple.  Let’s do one quick example of this.

Example.

Let’s let our matrix be

\left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right)

and now we consider our equation det(A - I\lambda) = 0, so we have that our matrix A - I\lambda is equal to

\left( \begin{array}{cc} (2 - \lambda) & 1 \\ 1 & (2 - \lambda) \end{array}\right)

so what’s the determinant of this?  Well, it’s just

 

(2 - \lambda)(2 - \lambda) - 1 = 4 - 4\lambda + \lambda^{2} - 1

 

and this is just the same as

 

\lambda^{2} - 4\lambda + 3 = (\lambda - 3)(\lambda - 1) = 0

 

This means that \lambda = \{1, 3\}.  Cool!

Now, to find the corresponding eigenvector, it’s easy.  Just plug in the eigenvalues to the equation.  So, for example, for the eigenvalue 3, we have:

\left(\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right) \left(\begin{array}{c} v_{1} \\ v_{2} \end{array}\right) = 3\left(\begin{array}{c} v_{1} \\ v_{2} \end{array}\right)

where we’ve replaced v with a matrix of the same form.  We multiply this out:

2v_{1} + v_{2} = 3v_{1}

v_{1} + 2v_{2} = 3v_{2}

and solving these equations, we find that v_{1} = v_{2}, so the eigenvector is any vector of the form

\left( \begin{array}{c} t \\ t \end{array}\right)

for any t\in {\mathbb R} except 0. 

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