I’m trying to figure out how to calculate the lines of invariance of a transformation using eigenvectors. I’m nearly there I just don’t understand how to work out whether it passes through the origin (y=mx) or not (y=mx+c).
I have the matrix $$ \begin{equation} M= \begin{bmatrix} 3 & 3 \\ 4 & 7 \\ \end{bmatrix} \end{equation} $$
And then using $\overrightarrow{v}$(M-$\lambda$I2)=0 I found the matrix: $$ \begin{bmatrix} 3-\lambda & 3 \\ 4 & 7-\lambda\ \\ \end{bmatrix} $$
I then calculated the eigenvalues by setting the determinant of this matrix equal to 0.
\begin{equation} \begin{vmatrix} 3-\lambda\ & 3 \\ 4 & 7-\lambda\ \\ \end{vmatrix} = \lambda^2-10\lambda + 9 = 0\ \end{equation}
Which gives $$ \lambda_1=9, \lambda_2=1 $$
Then substituting $\lambda$=1: $$ \begin{bmatrix} 2 & 3 \\ 4 & 6 \\ \end{bmatrix} $$
And setting up the matrix equation:
$$ \begin{equation} \begin{bmatrix} 2 & 3 \\ 4 & 6 \\ \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \end{equation} $$
Which gives the system of equations: $$ \left\{ \begin{array}{c} 2x + 3y = 0 \\ 4x + 6y = 0 \\ \end{array} \right. $$
Which both simplify to give the line equation y = -$\frac23$x
Following the same working, but with $\lambda$=9, yields the equation y=2x. I know from this example that this equation should be y=2x+c, but I don’t know how to know whether the line should have the “+c” or not from this method.
EDIT: I had this problem come up in class a few days ago, and derived the line equations y=-$\frac23$x and y=2x+c but using a different method. So after the transformation represented by the matrix is performed, all the points on the line y=-$\frac23$x remain in the same place (a line of invariant points). Any line of the form y=2x+c maps the points on that line to another point on the same line (a line of invariance). My question was, is there a way to determine whether or not a line is in the form y=mx or y=mx + c using this method, that is, does it pass through the origin or not.
One of the motivations for the concept of eigenvectors is the question “which lines are mapped to themselves?” The nonzero eigenvectors of a linear transformation give the directions of such lines that pass through the origin, but don’t directly say anything about lines that don’t pass through the origin. One could certainly come up with some ad-hoc procedure for finding these other invariant lines, but for $\mathbb R^2$ this can be turned into a pure eigenvalue problem by passing to homogeneous coordinates.
Writing the equation $ax+by+c=0$ of a line as $(a,b,c)(x,y,1)^T = \mathbf l^T\mathbf x=0$, we can see that lines in $\mathbb R^2$ can be represented by homogeneous coordinate vectors. Given an invertible $3\times3$ matrix $A$ that defines a point transformation $\mathbf x'=A\mathbf x$ we have $$\mathbf l^T\mathbf x = \mathbf l^T\left(A^{-1}\mathbf x'\right) = \left(A^{-T}\mathbf l\right)\mathbf x'=0,$$ i.e., lines transform as $\mathbf l' = A^{-T}\mathbf l$. Thus, the lines that are invariant under $A$ are the eigenvectors of $A^{-T}$. Since a matrix and its inverse have the same eigenvectors, finding the invariant lines of $A$ is a matter of computing its left eigenvectors.
For your specific problem, the homogeneous transformation matrix is $$A = \begin{bmatrix} M&\mathbf 0\\\mathbf 0^T&1\end{bmatrix} = \begin{bmatrix}3&3&0\\4&7&0\\0&0&1\end{bmatrix}$$ with eigenvalues $9$ and $1$. The left eigenspace of $9$ is spanned by $(2,3,0)^T$, which corresponds to the line with Cartesian equation $2x+3y=0$. The eigenspace of $1$, on the other hand, is two-dimensional. It is spanned by $(2,-1,0)^T$ and $(0,0,1)^T$, so every nonzero linear combination $\lambda(2,-1,0)+\mu(0,0,1)$ of these two vectors corresponds to an invariant line of $M$. When $\lambda=0$, we get the line at infinity for any value of $\mu$, so we can w.l.o.g. take $\lambda = 1$. Therefore, the family of lines with equations of the form $2x-y=c$, $c\in\mathbb R$ are the other invariant lines of $M$.
For an affine transformation, the last row of the corresponding $3\times3$ homogeneous matrix $A$ will be $(0,0,1)$, so it will always have $(0,0,1)^T$ as a left eigenvector with eigenvalue $1$. Geometrically, this means that the line at infinity is an invariant line of every affine transformation. On the other hand, when you have a linear transformation, the last column of $A$ is $(0,0,1)^T$, so any other linearly-independent left eigenvectors will have a final coordinate of $0$. This leads to a simple rule for finding invariant lines of a (invertible) linear transformation of $\mathbb R^2$ given by the $2\times2$ matrix $M$ that don’t pass through the origin: they are families of parallel lines with a common direction vector given by an eigenvector of $M^T$ with eigenvalue $1$.