We consider the
linear mapping
-
given by the matrix
-
![{\displaystyle {}M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b244cd53cb095060badd9641bcea5bb80a3253c)
The question whether this mapping has
eigenvalues,
leads to the question whether there exists some
,
such that the equation
-
![{\displaystyle {}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd9ac051863312f58bbadd81e47083f9a4dee6ff)
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 nullvector is not an eigenvector. Hence, suppose that
.
Both equations combined yield the condition
-
![{\displaystyle {}5y=\lambda x=\lambda ^{2}y\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/979ccad565746f4441f2816251dc572189e458f0)
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
-
![{\displaystyle {}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\sqrt {5}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3cd31c50d888dac119dc82b62108db9b4bc7a31)
We write this as
-
![{\displaystyle {}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}{\sqrt {5}}&0\\0&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3dcc01335a6e0828b2ba7f5eb404e9de7dee66eb)
and as
-
![{\displaystyle {}{\begin{pmatrix}{\sqrt {5}}&-5\\-1&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a354313e43e30f4ad1ad65009feb132af563d9b)
This system can be solved easily, the solution space has dimension one, and
-
![{\displaystyle {}v={\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/719c74f2667211e82a0654a5948163d595d42cfb)
is a basic solution.
For
,
we do the same steps, and the vector
-
![{\displaystyle {}w={\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f425a8c288d2320922a6c6e2bff8153bfd3497a)
is a basic solution. Thus over
, the numbers
and
are eigenvalues, and the corresponding
eigenspaces
are
-