Matrix/2x2/Shearing matrices/Eigenvalues and diagonalizable/Example

We consider ${\displaystyle {}2\times 2}$-shearing matrices

${\displaystyle {\begin{pmatrix}1&a\\0&1\end{pmatrix}}}$

with ${\displaystyle {}a\in K}$. The condition for some ${\displaystyle {}\lambda \in K}$ to be an eigenvalue means

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

This yields the equations

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

For ${\displaystyle {}\lambda \neq 1}$, we get ${\displaystyle {}y=0}$ and hence also ${\displaystyle {}x=0}$, that is, only ${\displaystyle {}1}$ can be an eigenvalue. In this case, the second equation is fulfilled, and the first equation becomes

${\displaystyle x+ay=x{\text{ and }}ay=0.}$

For ${\displaystyle {}a\neq 0}$, we get ${\displaystyle {}y=0}$ and thus ${\displaystyle {}{\begin{pmatrix}x\\0\end{pmatrix}}}$ is the eigenspace for the eigenvalue ${\displaystyle {}1}$, and ${\displaystyle {}{\begin{pmatrix}1\\0\end{pmatrix}}}$ is an eigenvector which spans this eigenspace. For ${\displaystyle {}a=0}$, we have the identity matrix, and the eigenspace for the eigenvalue ${\displaystyle {}1}$ is the total plane. For ${\displaystyle {}a\neq 0}$, there is a one-dimensional eigenspace, and the mapping is not diagonalizable.