# Matrix/Eigenvalues/0510/Q and R/Example

We consider the linear mapping
    $\varphi \colon \mathbb {Q} ^{2}\longrightarrow \mathbb {Q} ^{2},{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}5y\\x\end{pmatrix}},$
given by the matrix
    $M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}\,.$
The question whether this mapping has eigenvalues, leads to the question whether there exists some $\lambda \in \mathbb {Q}$, such that the equation
    ${\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,$
has a nontrivial solution $(x,y)\neq (0,0)$. For a given $\lambda$, 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“ $\lambda$, to a nonlinear problem. The system of equations above is
    $5y=\lambda x{\text{ and }}x=\lambda y.$
For $y=0$, we get $x=0$, but the null vector is not an eigenvector. Hence, suppose that $y\neq 0$. Both equations combined yield the condition
    $5y=\lambda x=\lambda ^{2}y\,,$
hence $5=\lambda ^{2}$. But in $\mathbb {Q}$, the number $5$ does not have a square root, therefore there is no solution, and that means that $\varphi$ has no eigenvalues and no eigenvectors.
Now we consider the matrix $M$ as a real matrix, and look at the corresponding mapping
    $\psi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2},{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}5y\\x\end{pmatrix}}.$
The same computations as above lead to the condition $5=\lambda ^{2}$, and within the real numbers, we have the two solutions
    $\lambda _{1}={\sqrt {5}}\,\,{\text{ and }}\,\,\lambda _{2}=-{\sqrt {5}}.$
For both values, we have now to find the eigenvectors. First, we consider the case $\lambda ={\sqrt {5}}$, which yields the linear system
    ${\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\sqrt {5}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.$
We write this as
    ${\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}{\sqrt {5}}&0\\0&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}\,$
and as
    ${\begin{pmatrix}{\sqrt {5}}&-5\\-1&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\,.$
This system can be solved easily, the solution space has dimension one, and
    $v={\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\,$
is a basic solution.
For $\lambda =-{\sqrt {5}}$, we do the same steps, and the vector
    $w={\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\,$
is a basic solution. Thus over $\mathbb {R}$, the numbers ${\sqrt {5}}$ and $-{\sqrt {5}}$ are eigenvalues, and the corresponding eigenspaces are
    $\operatorname {Eig} _{\sqrt {5}}{\left(\psi \right)}={\left\{s{\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}\,\,{\text{ and }}\,\,\operatorname {Eig} _{-{\sqrt {5}}}{\left(\psi \right)}={\left\{s{\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}.$