Search This Blog


Sunday, November 25, 2012

A Difficult Shear

Today's lesson is from Ch 9, Ex 15, Q14 David Rayner (

a) Find and draw the image of the square O (0, 0), P (1, 1), Q (0, 2), R (-1, 1) under the 
Square transformed to Parallelogram (Geogebra 4.0, www.

b) Show that the transformation is a shear and find the equation of the invariant line.

Mary says, from the drawing, it is evident that O’ = O and R’ = R. Hence the invariant line is y = -x.  If you need convincing, the gradient of the line is (-1 - 0)/(1 - 0)   = -1 and its y intercept is 0.  Next, look at the determinant: det M  = (4)(-2) – (3)(-3) = -8 + 9 = 1.  For IGCSE purposes, we can understand the determinant as an area scaling factor. Since the scale factor is unity, the image parallelogram must have the same area as the original square.   M is not a congruency, since a square is transformed into to a parallelogram.  There is just one line of invariant points, y = -x.  Finally, M preserves areas.   M must be a shear. 

Johnny says, but perhaps there is another invariant line? Actually, a little bit more algebra will prove this transformation has only one line of invariant points.  For an invariant point, P’ = P by definition.  That is, after the transformation, the point has not moved. That is exactly what we mean by invariant!!  Note that for any matrix transformation, the point O (0, 0) must always be an invariant point since

This is exactly the same as the equation that we found previously.  Thus we can say safely there is only one line of invariant points.

I summarise the lesson by reminding the students that, of the transformations we know, both reflections and shears have one invariant line. However, the matrix representing the reflection will have det Re = -1 whereas, as we have already seen, the shear will have det H = 1. 

It works.  Supposing that something exists, and then seeing where it leads us is quite characteristic of algebraic geometry.  Our motto is “Only a contradiction implies a contradiction.” That is, if our supposition does not lead us into a contradiction, it is valid. Equating elements: p =4, q = 3, k = -¾, s = ¼ . As the first matrix has the form of a shear, the transformation must be a shear.   

A parting question: Do you think the author, David Rayner, just played around with different values for a, b, c and d until he found a shear matrix?