Find the factor of the Shear T T represents a  transformation.
$$T = \begin{bmatrix}
3 & -2 \\
2 & -1 \\
\end{bmatrix}$$
i) Find the invariant points for the transformation T.
Which I found to be  $$y = x$$
ii) T is a transformation called a shear. The line of shear is the line of invariant
points for the shear. The factor of a shear gives the distance a point is moved
as a multiple of its perpendicular distance from the line of shear. What is the
factor of the shear T?
I applied the point (1,0) to the transformation, giving me (3,2). Then, I found the distance between the two points which is $2\sqrt2$.
However, I'm unsure on how to find the perpendicular distance.  Additionally, I don't know what "as a multiple of its perpendicular distance from the line of the shear" means.
I tried plotting the points on the graph, with the line y = x. Dropping a perpendicular from the line to the points, however, I'm unsure on how to proceed.
Any guidance would be appreciated.
 A: The distance from $(1,0)$ to the line $y=x$ is $\sqrt{2}/2$. The distance to its image is $2\sqrt{2}$ as you found out. Therefore the "factor of the shear" is $2\sqrt{2}\cdot 2/\sqrt{2}=4$.
A: $$T=\begin{pmatrix}  3 & -2 \\ 2 &-1 \end{pmatrix} \implies \det {T}=1. $$
Characteristic roots of $T$ are given by $\det|T- \lambda I|=0, \lambda= 1,1$
The eigen (invariant) vectors of $T$ are $$P_1= \begin{pmatrix}0 \\0 \end{pmatrix}.~~ P_2=\begin{pmatrix} 1 \\1 \end{pmatrix}.$$ Hence, $(0,0)$ and $(1,1)$ are the invariant points of $T$.
One does not know which figure (square, rectangle) you want to shear by $T$.
For instance the square with vertices  $(0,0),(1,0),(1,1),(0,1)$  can be sheared to a parallelogram with vertices $(0,0), (1,0), (1+k,1), (k,1)$ as below.
The transformation that shears a square  to a parallelogram parallel to $x$-axis is
$$T_1=\begin{pmatrix} 1 & k\\ 0 & 1 \end{pmatrix}, P_1=\begin{pmatrix} 0 \\ 0 \end{pmatrix}, P_2= \begin{pmatrix} 1 \\ 0\end{pmatrix}.$$ are invariant points, whereas the points $$\begin{pmatrix} 0 \\ 1 \end{pmatrix} \rightarrow \begin{pmatrix} k \\  1 \end{pmatrix}, ~~\begin{pmatrix} 1 \\ 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1+k \\ 1 \end{pmatrix}.$$
Finally, the measure of shear is $\theta=\cot^{-1} k$. In your case points $P_3, P_4$ are not given.
