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
-