Jump to content

Matrix/Eigenvalues/0510/Q and R/Example

From Wikiversity

We consider the linear mapping

given by the matrix

The question whether this mapping has eigenvalues, leads to the question whether there exists some , such that the equation

has a nontrivial solution . For a given , this is a linear problem and can be solved with the elimination algorithm. However, the question whether there exist eigenvalues at all, leads, due to the variable "eigenvalue parameter“ , to a nonlinear problem. The system of equations above is

For , we get , but the null vector is not an eigenvector. Hence, suppose that . Both equations combined yield the condition

hence . But in , the number does not have a square root, therefore there is no solution, and that means that has no eigenvalues and no eigenvectors.

Now we consider the matrix as a real matrix, and look at the corresponding mapping

The same computations as above lead to the condition , and within the real numbers, we have the two solutions

For both values, we have now to find the eigenvectors. First, we consider the case , which yields the linear system

We write this as

and as

This system can be solved easily, the solution space has dimension one, and

is a basic solution.

For , we do the same steps, and the vector

is a basic solution. Thus over , the numbers and are eigenvalues, and the corresponding eigenspaces are