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 nullvector 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

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