Have you ever wondered how software like photoshop can rotate, reflect and change the size of images? The software will use a matrix that represents the transformation required and apply it to your constraints to produce a transformation that you then see- this is as the transformations have set matrices that represent them. Read on to find out more about how this works!
Linear Transformations of Matrices Explanation
A linear transformation is a type of transformation with certain restrictions and factors placed on it. To be a linear transformation:
- The origin must always stay where it was before the transformation - it is an invariant point.
- Transformation must be linear - no powers of \(x\) or \(y\) can be included.
- Transformation must be able to be described by a matrix.
An invariant point or line is one that does not move during a linear transformation.
Considering these factors we can then experience several types of transformations and combinations of these. The linear transformations we can use matrices to represent are:
- Reflection
- Rotation
- Enlargement
- Stretches
Linear Transformations of Matrices Formula
When it comes to linear transformations there is a general formula that must be met for the matrix to represent a linear transformation. Any transformation must be in the form \(ax+by\). Consider the linear transformation \((T)\) of a point defined by the position vector \(\begin{bmatrix}x\\y\end{bmatrix}\). The resulting transformation could be written as this:\[T:\begin{bmatrix}x\\y\end{bmatrix}\rightarrow \begin{bmatrix}ax+by\\cx+dy\end{bmatrix}.\] Here we see \(ax+by\) and \(cx+dy\) to be describing the transformations in the \(x\) and \(y\) planes from the starting point to create our new point - the image (denoted by \(X'\) where \(X\) is the original vertex label). All we do is substitute in our values. Let's have a look at how this works.
We are going to find the coordinates of the image point with respect to the following transformation:\[T:\begin{bmatrix}3\\2\end{bmatrix} \rightarrow \begin{bmatrix}3x+2y\\5y\end{bmatrix}\]
While our bottom row here has the value of \(5y\) this could be written as \(0x+5y\) which is still in the form \(cx+dy\) so is a linear transformation.
Applying the transformation:\[\begin{align}T:\begin{bmatrix}3\\2\end{bmatrix} &\rightarrow \begin{bmatrix}3x+2y\\5y\end{bmatrix} \\&= \begin{bmatrix}3(3)+2(2)\\0(3)+5(2)\end{bmatrix}\\&=\begin{bmatrix}13\\10\end{bmatrix}\end{align}\]
This gives the coordinates of the image point to be \((13,10)\).
So we've seen how to apply a linear transformation but how do we denote these in matrix form? If we let the matrix be \(A\), we can write the matrix we need as thus:\[A\begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}ax+by\\cx+dy\end{bmatrix}\]Therefore, our linear transformation matrix is just a matrix of the coefficients of the linear transformation in the form we've already seen. This gives:\[\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}ax+by\\cx+dy\end{bmatrix}\]
Let's consider the transformation we saw above:\[T=\begin{bmatrix}3x+2y\\5y\end{bmatrix}\]We know the matrix is the coefficients of the transformation, so the matrix notation would read as such:\[A=\begin{bmatrix}3&2\\0&5\end{bmatrix}\]
Given the linear transformation matrix seen above, with a starting point of \((2,3)\) find the coordinates of the image point.
Solution:
\[\begin{align}&\begin{bmatrix}3&2\\0&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} \\= &\begin{bmatrix}3&2\\0&5\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix} \quad \mbox{apply matrix multiplication}\\=&\begin{bmatrix}(3\cdot3)+(2\cdot2)\\(0\cdot 3)+(5\cdot 2)\end{bmatrix}\\=&\begin{bmatrix}13\\10\end{bmatrix}\end{align}\]
As expected this yields the same result as before.
If this was a question with a shape with multiple vertices, we would apply the matrix to each vertex and its respective coordinates separately, then join the new vertices up to get our transformed shape.
Finding The Linear Transformations of Matrices
So now we've seen how to apply a linear transformation to a point using a matrix but how do we know what a matrix is actually describing in terms of transformation(s)? Luckily, in the \(2\times 2\) matrix form, every type of transformation has a set matrix for it to describe that transformation. Below you can see the conditions for each type of transformation and the associated matrix.
Reflection
For the reflections covered, every reflection has an invariant line located at the axis of reflection.
Reflection in the \(y\)-axis
Described by the matrix:\[A=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\]
Reflection in the \(x\)-axis
Described by the matrix:\[A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\]
Reflection about \(y=x\)
Described by the matrix:\[A=\begin{bmatrix}0&1\\1&0\end{bmatrix}\]
Reflection about \(y=-x\)
Described by the matrix:\[A=\begin{bmatrix}0&-1\\-1&0\end{bmatrix}\]
Rotation
Rotations have an invariant point at the origin and are rotated by \(\theta\) where a positive value indicates anticlockwise rotation. Rotations are described by the matrix:
\[A=\begin{bmatrix}\cos\theta &-\sin\theta \\\sin\theta &\cos\theta\end{bmatrix}\]
Enlargement and Stretches
A stretch or an enlargement is governed by the following matrix:\[A=\begin{bmatrix}a&0\\0&b\end{bmatrix}\]
The variable \(a\) is the scale factor for \(x\) and as such governs the stretch in the \(x\) direction in effect, it is the number we multiply the \(x\) coordinate by to achieve the new coordinate. For stretch in the \(y\) direction, \(b\) has the same effect on the \(y\) coordinate.
Enlargement
If we have \(a=b\), then the stretches in the \(x\) and \(y\) directions are the same and as such we have an enlargement, but if they are different then \(a\) governs change in \(x\) while \(b\) governs change in \(y\). There are no invariant lines in enlargement and the origin is the only invariant point.
Stretch parallel to \(x\)-axis
If we have a stretch parallel to the \(x\)-axis, the \(y\) position of any point will remain the same and is invariant. This type of transformation only stretches the \(x\) coordinates. It is governed by the matrix:\[A=\begin{bmatrix}a&0\\0&1\end{bmatrix}\]
Stretch parallel to \(y\) axis
If we have a stretch parallel to the \(y\)-axis, the \(x\) position of any point will remain the same and is invariant. This type of transformation only stretches the \(y\) coordinates. It is governed by the matrix:\[A=\begin{bmatrix}1&0\\0&b\end{bmatrix}\]
Area change
With stretches and enlargements, we are changing the size of any shape made up of the vertices we are moving. As such, unlike with reflections and rotations, the area of the shape will change. Luckily we have a way to measure this change in the area. If matrix \(A\) represents the transformation, then \(\det{A}\) will give you the scale factor for the change in the area. For information on how to calculate the determinant see our article on Matrix Determinants.
If \(\det{A}\) is negative, the shape has been reflected
Successive Transformations
So now you should have all the information you need to find the linear transformation being described by a matrix, but what if there have been multiple transformations applied? Successive transformations can be described in one matrix by multiplying them together. Let's have a recap of matrix multiplication so you can perform successive transformations. Assumed below that \(A\) and \(B\) are two separate transformations with \(A\) being the first one to happen operationally and \(AB\) is the matrix describing the successive transformations.\[A=\begin{bmatrix}a_1&a_2\\a_3&a_4\end{bmatrix}\quad B=\begin{bmatrix}b_1&b_2\\b_3&b_4\end{bmatrix}\]\[\begin{align}AB&=\begin{bmatrix}a_1&a_2\\a_3&a_4\end{bmatrix}\begin{bmatrix}b_1&b_2\\b_3&b_4\end{bmatrix}\\&=\begin{bmatrix}(a_1b_1+a_2b_3)&(a_1b_2+a_2b_4)\\(a_3b_1+a_4b_3)&(a_3b_2+a_4b_4)\end{bmatrix}\end{align}\]
A point goes through a reflection in the \(y\) axis followed by a rotation of \(90^\circ\) anticlockwise. What is the matrix that describes these successive transformations?
Solution:
Let \(A\) be a reflection in the \(y\) axis:\[A=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\]
Let \(B\) be a rotation of \(90^\circ\) anticlockwise:\[\begin{align} B&=\begin{bmatrix}\cos\theta &-\sin\theta \\\sin\theta &\cos\theta\end{bmatrix}\\&=\begin{bmatrix}\cos(90) &-\sin(90) \\\sin(90) &\cos(90)\end{bmatrix}\\&= \begin{bmatrix}0 &-1 \\1 &0\end{bmatrix} \end{align}\]
Let \(AB\) be the successive transformations of \(A\) then \(B\):\[\begin{align}AB&=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\begin{bmatrix}0 &-1 \\1 &0\end{bmatrix}\\&=\begin{bmatrix}(-1\cdot 0+0\cdot 1)&(-1\cdot -1+0\cdot 0)\\(0\cdot 0+1\cdot 1)&(0\cdot -1+1\cdot 0)\end{bmatrix}\\&=\begin{bmatrix}0&1\\1&0\end{bmatrix}\end{align}\]
Linear Transformations of 2×2 Matrices Examples
Let's now have a look at a few examples involving shapes and linear transformations. To start let's look at an enlargement/stretch transformation.
A triangle has vertices located at \(X=(0,3)\), \(Y=(2,4)\) and \(Z=(5,2)\). What are the new coordinates of the image shape, and what is the area scale factor if the transformation is governed by the matrix \(A\).\[A=\begin{bmatrix}1.2&0\\0&-2\end{bmatrix}\]
Solution:
Let's start by finding the new image coordinates.
\(X\) coordinate:
\[\begin{align}X'&=AX\\&=\begin{bmatrix}1.2&0\\0&-2\end{bmatrix}\begin{bmatrix}0\\3\end{bmatrix}\\&=\begin{bmatrix}1.2\cdot 0+0\cdot 3\\0\cdot 0+-2\cdot 3\end{bmatrix}\\&=\begin{bmatrix}0\\-6\end{bmatrix}\end{align}\]
Therefore the image of point \(X\) is located at \(X'=(0,-6)\)
\(Y\) coordinate:\[\begin{align}Y'&=AY\\&=\begin{bmatrix}1.2&0\\0&-2\end{bmatrix}\begin{bmatrix}2\\4\end{bmatrix}\\&=\begin{bmatrix}1.2\cdot 2+0\cdot 4\\0\cdot 2+-2\cdot 4\end{bmatrix}\\&=\begin{bmatrix}2.4\\-8\end{bmatrix}\end{align}\]
Therefore the image of point \(Y\) is located at \(Y'=(2.4,-8)\).
\(Z\) coordinate:
\[\begin{align}Z'&=AZ\\&=\begin{bmatrix}1.2&0\\0&-2\end{bmatrix}\begin{bmatrix}5\\2\end{bmatrix}\\&=\begin{bmatrix}1.2\cdot 5+0\cdot 2\\0\cdot 5+-2\cdot 2\end{bmatrix}\\&=\begin{bmatrix}6\\-4\end{bmatrix}\end{align}\]
Therefore the image of point \(Z\) is located at \(Z'=(6,-4)\).
Area scale factor:
\[\begin{align}\det{A}&=(1.2\cdot -2) - (0\cdot 0)\\&=-2.4\end{align}\]
In the image below we can see a sketch of the original triangle and the image after the transformation. Additionally, we can see the reflection we would expect with a negative determinant- the stretch has also reflected the triangle around the \(x\) axis.
Fig. 1. Sketch of the image after transformation.
Let's now look at an example of successive transformations when applied to a shape.
A triangle goes through a reflection in the \(y\) axis followed by a rotation of \(90^\circ\) anticlockwise. Said triangle has vertices located at \(X=(0,3)\), \(Y=(2,4)\) and \(Z=(5,2)\). Find the image coordinates and sketch.
Solution:
As seen previously:
Let \(A\) be a reflection in the \(y\) axis:\[A=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\]
Let \(B\) be a rotation of \(90^\circ\) anticlockwise:\[\begin{align} B&=\begin{bmatrix}0 &-1 \\1 &0\end{bmatrix} \end{align}\]
Let \(AB\) be the successive transformations of \(A\) then \(B\):\[\begin{align}AB&=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\begin{bmatrix}0 &-1 \\1 &0\end{bmatrix}\\&=\begin{bmatrix}0&1\\1&0\end{bmatrix}\end{align}\]
You may notice here that this successive transformation is the same as a reflection in \(y=x\) - you can see this in the image below.
\(X\) coordinate:
\[\begin{align}X'&=AX\\&=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0\\3\end{bmatrix}\\&=\begin{bmatrix}0\cdot 0+1\cdot 3\\1\cdot 0+0\cdot 3\end{bmatrix}\\&=\begin{bmatrix}3\\0\end{bmatrix}\end{align}\]
Therefore the image of point \(X\) is located at \(X'=(3,0)\).
\(Y\) coordinate:\[\begin{align}Y'&=AY\\&=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}2\\4\end{bmatrix}\\&=\begin{bmatrix}0\cdot 2+1\cdot 4\\1\cdot 2+0\cdot 4\end{bmatrix}\\&=\begin{bmatrix}4\\2\end{bmatrix}\end{align}\]
Therefore the image of point \(Y\) is located at \(Y'=(4,2)\).
\(Z\) coordinate:
\[\begin{align}Z'&=AZ\\&=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}5\\2\end{bmatrix}\\&=\begin{bmatrix}0\cdot 5+1\cdot 2\\1\cdot 5+0\cdot 2\end{bmatrix}\\&=\begin{bmatrix}2\\5\end{bmatrix}\end{align}\]
Therefore the image of point \(Z\) is located at \(Z'=(2,5)\).
Fig. 2. Sketch of the image, also showing how the successive transformations are equivalent to a reflection in \(y=x\).
Linear Transformations of 3×3 Matrices Examples
When applying linear transformations to a \(3\times 3\) matrix we are operating in the world of 3D transformations. These are more complicated than what we've looked at so far. Look for our article on Matrix Transformations in 3D for a full explanation and examples.
Linear Transformations of Matrices - Key takeaways
- A linear transformation has an invariant point at the origin and takes the form \[\begin{bmatrix}ax+by\\cx+dy\end{bmatrix}\]
- A linear transformation can be represented as a matrix of coefficients. This takes the form:\[\begin{bmatrix}a&b\\c&d\end{bmatrix}\]
- A reflection is governed by a matrix of \(0's\) and \(1's\) to represent reflection around an axis and this axis is an invariant line.
- A rotation is positive if anticlockwise and is governed by: \[\begin{bmatrix}\cos\theta &-\sin\theta \\\sin\theta &\cos\theta\end{bmatrix}\]
- A stretch or enlargement changes the size of a shape, with the determinant of the transformation matrix being the scale factor for the change in area and is governed by:\[\begin{bmatrix}a&0\\0&b\end{bmatrix}\]
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