TITLE: Checking if a vector can be eigenvector of an eigenvalue.
QUESTION [0 upvotes]: An affine transformation is given by:
$$\begin{bmatrix}
    3     & 4 & 8  \\
    -2   & -3 & 7  \\
   0  & 0 & 1 
\end{bmatrix}$$
The matrix
$$\begin{bmatrix}
    3     & 4   \\
    -2   & -3   \\ 
\end{bmatrix}$$
Has the eigenvalues $1,-1$. And the eigenvectors $(-2,1),(-1,1).$
I want to find the points $P$ so that $\overrightarrow{Pf(P)}$ is an eigenvector. I have calculated $\overrightarrow{Pf(P)}=(2x+4y+8,-2x-4y+7)$. Now I think I should see if this vector can be eigenvector of the eigenvalue $-1$ or $1$. How can I do that?

REPLY [1 votes]: Picking up where you left off, you have an explicit expression for $\overrightarrow{Pf(P)} = f(P)-P$ and two linearly independent eigenvectors $\mathbf v_1=(-2,1)^T$ and $\mathbf v_2=(-1,1)^T$. Recalling that any nonzero scalar multiple of an eigenvector is also an eigenvector, this means that we must have $f(P)-P=\alpha \mathbf v_1$ or $f(P)-P=\beta \mathbf v_2$ for some non-zero scalars $\alpha$ and $\beta$. Expanding the first equation produces the system $$\begin{align} 2x+4y+8 &= -2\alpha \\ -2x-4y+7 &= \alpha.\end{align}$$ Adding twice the second equation and dividing out a common factor of $2$ gives the equation $x+2y-11=0$. This line has $\mathbf v_1$ as a direction vector and, as you can verify for yourself, is mapped to itself by the transformation (I’ll come back to this point later). For $\mathbf v_2$ the corresponding system of equations is $$\begin{align} 2x+4y+8 &= -\beta \\ -2x-4y+7 &= \beta.\end{align}$$ This system is clearly inconsistent, so there are no points $P$ for which $f(P)-P$ lies in the span of $v_2$.  
There’s a slightly different approach to this problem that highlights the geometric meaning of the condition to be satisfied. Write $f(P)=L(P)+\mathbf b$, where $L$ is the linear part of $f$, and let $\mathbf v$ be an eigenvector of $L$ with eigenvalue $\lambda$. If $P$ satisfies $f(P)-P=\alpha\mathbf v$ for some $\alpha\ne0$, then $$\begin{align} f(P+t\mathbf v) - (P+t\mathbf v) &= L(P+t\mathbf v)+\mathbf b - P-t\mathbf v \\ &= L(P) + \lambda t\mathbf v + \mathbf b - P - t\mathbf v \\ &= f(P) - P + (\lambda-1) t\mathbf v \\ &= \alpha\mathbf v +(\lambda-1)t\mathbf v, \end{align}$$ which is also a scalar multiple of $\mathbf v$, so almost every point on the line $P+t\mathbf v$ also satisfies the condition. (If $P+t\mathbf v$ happens to be a fixed point of $f$, then technically it fails because $0$ is never an eigenvector.) However, the image of a point on this line is itself of the form $P+t\mathbf v$ and so lies on the line, therefore the line is mapped to itself by $f$. Hence, we are really looking for the invariant lines of $f$.  
Working in homogeneous coordinates, if $M$ is the matrix that represents $f$, then the invariant lines of $f$ are the eigenvectors of $$M^{-T} = \begin{bmatrix}3&-2&0\\4&-3&0\\-52&37&1\end{bmatrix}.$$ By inspection, $(0,0,1)^T$ is an eigenvector with eigenvalue $1$. This is the line at infinity, which is always mapped to itself by affine transformations. The remaining eigenvalues are those of the upper-right $2\times2$ submatrix, and are easily found to be $\pm1$. There are no other linearly independent eigenvectors of $1$, consistent with the calculation at top, and the usual methods produce $(1,2,-11)^T$ as an eigenvector of $-1$, which corresponds to the line $x+2y-11=0$, as the computation at the top also found.
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