Find the equation of a plane that is invariant with respect to the following transformation: \begin{pmatrix}4&-23&17\\ \:11&-43&30\\ \:15&-54&37\end{pmatrix}
Actually, I don't know the algorithm but tried to find it as follows: $$\begin{pmatrix}4&-23&17\\ \:11&-43&30\\ \:15&-54&37\end{pmatrix}\begin{pmatrix}x \\ y \\z\end{pmatrix}=\begin{pmatrix}x \\ y \\z\end{pmatrix}$$
And by simplifying this we get
$$\begin{pmatrix}15&-54&36\\ 0&-\frac{61}{5}&\frac{49}{5}\\ 0&0&\frac{4}{61}\end{pmatrix}\begin{pmatrix}x \\ y \\z\end{pmatrix}=0$$ Which just implies that the equation has only trivial solutions. Could anyone please tell the right track?
$\textbf{Answer:}$ $3x-3y+z=0$
Assuming that this invariant plane passes through the origin, its equation is of the form $\mathbf n\cdot \mathbf x=0$, where $\mathbf n$ is some vector normal to the plane. If $\mathbf x'=M\mathbf x$, where $M$ is the matrix in your problem, then for $$\mathbf n\cdot\mathbf x' = \mathbf n\cdot M\mathbf x = M^T\mathbf n\cdot\mathbf x = 0$$ to represent the same plane, we must have $M^T\mathbf n = \lambda\mathbf n$ for some nonzero real $\lambda$, i.e., $\mathbf n$ is an eigenvector of $M^T$ with a real eigenvalue.
It turns out that even if we consider planes that don’t pass through the origin, we get the same solution, so we didn’t lose anything by restricting our attention to planes through the origin.