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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.

ymxzxz
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  • It is not clear as to the vertices of which figure (square etc) you are shearing with this $T$.. You may see my partial answer below. – Z Ahmed Sep 06 '20 at 16:54

2 Answers2

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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 $2\sqrt{2}\cdot 2/\sqrt{2}=4$.

GReyes
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  • How did you find the distance of the point to the line? Additionally, what does "factor of the shear" actually mean? – ymxzxz Sep 06 '20 at 12:26
  • @ymxzxz The factor is the ratio of the distance of the input (point before the shear) to the line to the distance of the output (point after the shear) to the line. – Ben Grossmann Sep 06 '20 at 14:17
  • @y To find the distance from $x$ to the line spanned by the unit vector $v$, find the length of the orthogonal component. That is, the distance is equal to $|x - (v^Tx),v|$. – Ben Grossmann Sep 06 '20 at 14:18
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$$T=\begin{pmatrix} 3 & -2 \\ 2 &-1 \end{pmatrix} \implies \det {T}=1. $$ 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.

Z Ahmed
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