I have the matrix
$$Q=\begin{bmatrix}-1&2\\0&1\\\end{bmatrix}$$
and want to find the invariant points.
To do this, I solve the equation:
$$\begin{bmatrix}-1&2\\0&1\\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix} =\begin{bmatrix}x\\y\\\end{bmatrix} $$
The first equation is $-x + 2y = x \Rightarrow y = x$. This shows that any points on $y=x$ are invariant under the transformation represented by $Q$.
The second equation, however, gives $y = y$ which encodes nothing? Just interested in the origin of this degeneracy and whether it has any significance…
Further question prompted by an answer:
Sometimes when I solve the matrix equation, the first and second equation both give the condition. In this case only the first equation gives $y=x$ and the second equation is degenerate. In what cases would I have both equations leading to the same condition or one being degenerate?
Best Answer
$Q$ has rank $2$ and eigenvalues $+1$ and $-1$. The invariant points correspond to the eigenvectors with eigenvalue $+1$. As you have found, these are scalar multiples of $\begin{bmatrix}1\\1\\\end{bmatrix}$ and so lie on the line $y=x$.
The eigenvectors with eigenvalue $-1$ are scalar multiples of $\begin{bmatrix}1\\0\\\end{bmatrix}$ and so lie on the line $y=0$. In other words, points on the $x$ axis are reflected in the $y$ axis.
If $\lambda$ is an eigenvalue of $Q$ then $\det(Q-\lambda I) = 0$, in other words $Q-\lambda I$ is degenerate. So when you solve $(Q - \lambda I)X=0$ to find the eigenvector for a given eigenvalue, the final row of $(Q-\lambda I)X$ is always redundant.