Matrix/Eigenvalues/0510/Q and R/Example

We consider the linear mapping

${\displaystyle \varphi \colon \mathbb {Q} ^{2}\longrightarrow \mathbb {Q} ^{2},{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}5y\\x\end{pmatrix}},}$

given by the matrix

${\displaystyle {}M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}\,.}$

The question whether this mapping has eigenvalues, leads to the question whether there exists some ${\displaystyle {}\lambda \in \mathbb {Q} }$, such that the equation

${\displaystyle {}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,}$

has a nontrivial solution ${\displaystyle {}(x,y)\neq (0,0)}$. For a given ${\displaystyle {}\lambda }$, 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“ ${\displaystyle {}\lambda }$, to a nonlinear problem. The system of equations above is

${\displaystyle 5y=\lambda x{\text{ and }}x=\lambda y.}$

For ${\displaystyle {}y=0}$, we get ${\displaystyle {}x=0}$, but the null vector is not an eigenvector. Hence, suppose that ${\displaystyle {}y\neq 0}$. Both equations combined yield the condition

${\displaystyle {}5y=\lambda x=\lambda ^{2}y\,,}$

hence ${\displaystyle {}5=\lambda ^{2}}$. But in ${\displaystyle {}\mathbb {Q} }$, the number ${\displaystyle {}5}$ does not have a square root, therefore there is no solution, and that means that ${\displaystyle {}\varphi }$ has no eigenvalues and no eigenvectors.

Now we consider the matrix ${\displaystyle {}M}$ as a real matrix, and look at the corresponding mapping

${\displaystyle \psi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2},{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}5y\\x\end{pmatrix}}.}$

The same computations as above lead to the condition ${\displaystyle {}5=\lambda ^{2}}$, and within the real numbers, we have the two solutions

${\displaystyle \lambda _{1}={\sqrt {5}}\,\,{\text{ and }}\,\,\lambda _{2}=-{\sqrt {5}}.}$

For both values, we have now to find the eigenvectors. First, we consider the case ${\displaystyle {}\lambda ={\sqrt {5}}}$, which yields the linear system

${\displaystyle {}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\sqrt {5}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.}$

We write this as

${\displaystyle {}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}{\sqrt {5}}&0\\0&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}\,}$

and as

${\displaystyle {}{\begin{pmatrix}{\sqrt {5}}&-5\\-1&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\,.}$

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

${\displaystyle {}v={\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\,}$

is a basic solution.

For ${\displaystyle {}\lambda =-{\sqrt {5}}}$, we do the same steps, and the vector

${\displaystyle {}w={\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\,}$

is a basic solution. Thus over ${\displaystyle {}\mathbb {R} }$, the numbers ${\displaystyle {}{\sqrt {5}}}$ and ${\displaystyle {}-{\sqrt {5}}}$ are eigenvalues, and the corresponding eigenspaces are

${\displaystyle \operatorname {Eig} _{\sqrt {5}}{\left(\psi \right)}={\left\{s{\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}\,\,{\text{ and }}\,\,\operatorname {Eig} _{-{\sqrt {5}}}{\left(\psi \right)}={\left\{s{\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}.}$