# Find invariant points under matrix transformation The matrix: $$Q=\begin{bmatrix}-1&2\\0&1\\\end{bmatrix}$$

Find invariant points under matrix transformation
The matrix:
$Q=\left[\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right]$
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Nevaeh Jensen
Step 1
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
$\left[\begin{array}{c}1\\ 1\end{array}\right]$
and so lie on the line $y=x.$
The eigenvectors with eigenvalue -1 are scalar multiples of
$\left[\begin{array}{c}1\\ 0\end{array}\right]$
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 $det\left(Q-\lambda I\right)=0$, in other words $Q-\lambda I$ is degenerate. So when you solve
$\left(Q-\lambda I\right)X=0$ to find the eigenvector for a given eigenvalue, the final row of $\left(Q-\lambda I\right)X$ is always redundant.

Jason Duke
Step 1
The non-zero invariant points are eigenvectors with eigenvalue 1. If the problem has full rank, then the only solution would be the zero vector.
As you have shown, the only condition is $x=y$. That is, the invariant points are multiples of
$\left[\begin{array}{c}1\\ 1\end{array}\right].$