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We've seen that affine transformations can, in general, squash and stretch things out of shape, (although straight lines remain straight). We've also looked at a special class of affine transformations called isometries, which preserve lengths and which, as a consequence, preserve shape and size.
But there is also an in-between case. There are transformations that do preserve shape but don't preserve size. However, we don't define them as preserving "shape", because that's a bit woolly. We come at it in a slightly different way. Recall that, in general, any linear transformation (and hence also any affine transformation) preserves the ratio of lengths of parallel segments, but not of non-parallel segments. We now define a similarity transformation to be an affine transformation that preserves the ratio of lengths of all segments. To the right is an example of a similarity transformation. This particular one shrinks things towards P and then rotates them 60° about P. |
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Noting that a similarity transformation preserves ratios of segments, consider any two segments [AB] and [CD]. We know that
But that can be rearranged as:
That is, the image of any segment compared to the original segment is always the same. i.e., if the image of [AB] is twice as long as [AB], then the image of any other segment will be twice as long as that segment. In other words, all segments are growing or shrinking by the same factor. This factor is called the scale factor of the similarity transformation, and usually denoted by the letter k.
In the example at the top of this page, k = 1/2.
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To show this, we use the cosine rule. Suppose you're given an angle ABC. Then consider the triangle ABC. The cosine rule tells you that:
But if you apply the transformation and do the cosine rule in the image triangle you get:
You may recall from Junior Cert that triangles whose angles are the same are called equiangular triangles or similar triangles. (Remember there was a theorem that in equiangular triangles, the corresponding sides are proportional.) We've just shown that similarity transformations preserve angle measure. It therefore follows that they map triangles to similar triangles (hence the name). |
Exercise:
Using the formula ½abSinC for the area of a triangle, show that, under a similarity transformation, the area of the image of a triangle is k2 times the area of the original triangle.
By the same line of reasoning that we used for isometries, this is also true of other shapes.
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Interestingly, the Leaving Certificate Ordinary Level course includes the study of one particular type of similarity transformation, called an enlargement. It's not dealt with from the perspective we've taken, obviously, but students are expected to be able to construct images under it and to understand and use the ratio results, including the area one.
To construct the image, you draw rays (half-lines) out from the centre of enlargement through the object point. You measure that distance and multiply by the scale factor to figure out how far out along the ray is the image point. The diagram on the right shows an enlargement, centred at P and of scale factor k. When k<1, it's sometimes called a reduction, but it's not incorrect to still call it an enlargement. Another name for an enlargement is a dilation. Although not the term used on the Leaving Cert syllabus, this is the term you'll see used most often on the internet and in dynamic geometry software packages. The transformation matrix for an enlargement centred at the origin is particularly straightforward. It's just:
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Enlargement (dilation)
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We can follow the same sort of approach here as we did with isometries:
Suppose you're just given an affine transformation. That is, a transformation of the form:
How do you figure out whether or not it is a similarity transformation? As before, it's clear that the translation part (the k1 and the k2) isn't going to affect whether or not it's a similarity transformation, since that bit will always preserve lengths (and therefore certainly preserve ratios of lengths). So the question is, can you tell from the values of a, b, c, and d whether this is a similarity transformation? The answer is again "yes".
Again, we can make things a bit easier by considering only distances from the origin to a point. This is justified by the same line of reasoning as wih isometries (any linear transformation will multiply parallel segments by the same factor, so any segment will be treated the same as a parallel segment from the origin).
My question has now become: what conditions on a, b, c, and d will result in the distance from (0, 0) to (x', y') always being equal to a fixed multiple k of the distance from (0, 0) to (x, y)? Time for the donkey-work again. Again, I'll use distances squared instead of distances, to save writing out the square roots. (But this means I want the squared distances to go by a factor of k2.)
Precisely as before, the squared distance from (0, 0) to (x, y) is x2 + y2.
Also as before, x' = ax + by and y' = cx + dy, so the squared distance from (0, 0) to (x', y') is (ax + by)2 + (ax + by)2.
And yet again as before, multiplying this out and gathering appropriately gives: (a2+c2)x2 + (b2+d2)y2 + 2(ab+cd)xy.
So, for a similarity transformation, we want the squared distance from (0, 0) to (x', y') to be k2 times the squared distance from (0, 0) to (x, y).
That is, we want k2x2 + k2y2 = (a2+c2)x2 + (b2+d2)y2 + 2(ab+cd)xy.
But remember that again we want this to be true for all x and y.
This will happen if and only if the coefficients of x2, y2, and xy match on the two sides.
That is, it will happen if and only if: (1) a2+c2 = k2, (2) b2+d2 = k2, and (3) ab+cd = 0.
So, the above three conditions on a, b, c, and d are the conditions for a similarity transformation.
Again, I don't think you need to know the above three conditions off by heart, but I think you probably could be asked to establish them, (i.e. more or less as above).
If we divide the equations above across by k2, the right-hand side of the first two becomes 1, and you should be able to see that a/k, b/k, c/k and d/k are now in exactly the same position as were a, b, c, and d in the case of the isometry.
Hence, the matrix for a similarity transformation is k times the matrix of an isometry. Another way of looking at this is that, noting what the matrix of an enlargement looks like (above), every similarity transformation is the composition of an enlargement and an isometry (with an appropriate tweak to the translation part).
That is, every similarity transformation is of the form:
Let's stop a moment to take stock:
| Transformation type | Some salient properties | Includes |
|---|---|---|
| non-linear, non-affine | might turn lines into curves (or into segments or other subsets of lines) | The weird one on the first page; (that one's actually called "circular inversion") |
| affine | lines stay straight | transformations that don't necessarily leave triangles the same shape. In fact, given any two triangles, it's always possible to find an affine transformation that maps one to the other. (We didn't prove that, by the way.) |
| linear | affine transf. that leaves the origin fixed, (i.e., has no translation bit.) | |
| isometry | preserves length, shape, size, area, congruence. | translations, rotations, reflections, glide reflections |
| similarity | preserves shape but not size; has a (fixed) scale factor. | enlargement; enlargement followed by any isometry |
This is a term that I could have introduced earlier, as we have been dealing with the concept all along. "Invariant" simply means "doesn't change". We have already established the invariance or non-invariance of lots of things under various different types of transformattion. For example "length" is invariant under an isometry, but is not, in general, invariant under a linear or affine transformation.
Something that's invariant under an affine transformation is called an "affine invariant". Thus, for example, we say things like "Parallelism is an affine invariant", "area is not an affine invariant", etc.
We have already established (or can establish) the invariances in the table below. Now, don't go learning these off by heart; all you need is to have a mental picture of each of these types of transformation, and you will automatically know all of them.
(I'm including a column for "all invertible transformations, even though we don't study non-affine transformations, just to remind you that all of these invariances that we have established are significant.)
| Property | All invertible transformations | Affine | Similarity | Isometry |
|---|---|---|---|---|
| Being a point | Yes | Yes | Yes | Yes |
| Being a line | No | Yes | Yes | Yes |
| Being a triangle | No | Yes | Yes | Yes |
| Parallelisim | No | Yes | Yes | Yes |
| Length | No | No | No | Yes |
| Ratio of lengths of parallel segments | No | Yes | Yes | Yes |
| Hence, ratio of segments on a single line | No | Yes | Yes | Yes |
| Hence, "midpointism" | No | Yes | Yes | Yes |
| Ratio of all segments | No | No | Yes | Yes |
| Angle measure | No | No | Yes | Yes |
| Hence, perpendicularity | No | No | Yes | Yes |
| Area | No | No | No | Yes | Ratio of areas* (see ex. below) | No | Yes* | Yes | Yes |
| Orientation | No | No | No | No |
| Tangency* (see below) | ? | Yes | Yes | Yes |
| Being a circle | No | No | Yes | Yes |
Exercise:
(i) (Algebra a bit messy!) Show that under any affine transformation the area of the triangle with vertices at (0, 0), (x1, y1), (x2, y2) gets multiplied by the determinant of the transformation matrix.
(ii) Write out an argument why the same will hold for all areas (similar idea as we used before).
(iii) Deduce that ratio of areas is an affine invariant.
One or two other things are worth mentioning. Firstly, "being a point" is an invariant under any transformation. This is arguably a trivial statement, but it's just to remind you of what a transformation is: it's a function that sends each point in the plane to a point in the plane.
Secondly, again almost trivially, if a point is a member of a set of points (such as a line), then, by definition, its image is a member of the image set. It couldn't be otherwise, since the image of a set of points is defined to be the set of all images of the points in the original set. A consequence of this is that the point(s) of intersection of two sets (e.g. point of intersection of two lines) will map to the point(s) of intersection of the images.
Also, with any invertible transformation, two distinct points cannot map onto a single point. (Also, of course, one point cannot turn into two points.) A consequence of this is that the number of points of intersection of two objects is invariant under any invertible transformation.
A further consequence of this is that tangency is an affine invariant. Any line and any circle always intersect in 0, 1 or 2 points. When a line and a circle intersect in precisely one point, the line is said to be a tangent to the circle at that point. Under an affine transformation, the image of the line intersects the image of the circle at exactly one point also. The image of the circle, as it happens, will be a circle or an ellipse, and the image line, since it shares exactly one point with it, will be a tangent to it.
(In case you're wondering why I put a question mark for tangency in the case of an invertible transformation in general, it's because, not having any idea what kinds of object the images of lines and circles might be in such a case, the notion of tangency might not even be meaningful.)
Exercise:
(i) Which of the following are always true under an affine transformation:
Reminder:The centroid is the point of intersaction of the medians, (a median being a line joining a vertex to the midpoint of the opposite side). The incentre is the point of intersection of the angle bisectors. The circumcentre is the point of intersection of the perpendicular bisectors of the sides. The orthocentre is the point of intersection of the altitudes, (an altitude being a line through a vertex perpendicular to the opposite side - i.e., a "perpendicular height" line).
(ii) Which of the above are true under a similarity transformation?
