Given the matrix $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ with integer coefficients, rational eigenvalue, and determinant $1$, show $A$ acts as a shearing along its eigenvector.
Here is what I have so far: from the characteristic polynomial, we have $$\lambda^2 - (a+d)\lambda + 1 = 0$$
and by the quadratic formula, we get $$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4}}{2}.$$
Now we assumed that $\lambda$ is rational, implying the square root must be a rational. But $a, d \in \Bbb{Z} \Longrightarrow \mbox{the square root is either integer or irrational}$. It is the former case when $a+d = 2$. It is also possible that $a+d = -2$, but that would make $det(A) = -1$, a contradiction.
Continuing, we get the eigenvalue is $1$. Alternatively, we substitute $2$ into our C.P. to get $$\lambda^2 - 2\lambda + 1 = \left(\lambda - 1\right)^2$$ implying the algebraic multiplicity of the eigenvalue $1$ is $2$.
Now finding the eigenvector, we find the null space of $A - \lambda I$. $$A - I =\begin{pmatrix}a-1&b&0\\c&d-1&0\end{pmatrix} \Longrightarrow \begin{pmatrix}a-1&b&0\\0&0&0\end{pmatrix}$$
and our only eigenvector is $$\vec{e}_1 = x_2\begin{pmatrix}-b\\a-1\end{pmatrix}$$
implying its geometric multiplicity is $1$.
Now I would like some hints on showing that $A$ is a shearing along its eigenvector. Do I have to show $A\vec{e}_1 = \begin{pmatrix}1&k\\0&1\end{pmatrix}$? I am unclear as what to do at this point.
As an example take the matrix $$\begin{pmatrix}6&5\\-5&-4\end{pmatrix}$$
Then we have as its eigenvector $$x_2\begin{pmatrix}-1\\1\end{pmatrix};\hspace{5 pt} \mbox{$x_2 = 5$}$$
But I don't know what to do now.
OK, so you've found the eigenvector. Now you want to find a vector $e_2$ so that $Ae_2 = e_1+e_2$. Then the matrix of your linear transformation with respect to the basis $\{e_1,e_2\}$ will be $$\begin{bmatrix}1&1\\0&1\end{bmatrix}\,.$$ This represents a shear because you're keeping $e_2$ at the same "height" relative to $e_1$, but sliding it along the $e_1$ direction. The terminology comes from geology :)
How are you guaranteed the existence of such a vector $e_2$? You need to know that $e_1$ is in the image (column space) of $A-I$. And this happens because $(A-I)^2 = O$. Why?