Linear transformations
Master linear transformations in Year 11 VCE Specialist Mathematics. A linear transformation moves every point of the plane using a single two-by-two matrix, and the columns of that matrix are simply the images of the two basis directions. It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Transformations of the plane topic of Unit 2.
You will learn to find the image of a point or a whole polygon, build the matrix from where the basis vectors go, use the linearity rules, and see why the origin stays fixed and how the determinant scales area — the groundwork for reflections, rotations and dilations later in the course.
Theory
A linear transformation of the plane sends each point \((x,y)\) to an image \((x',y')\) using a \(2\times2\) matrix, written \(T(\mathbf{x})=A\mathbf{x}\). In Year 11 Specialist Mathematics the columns of \(A\) are the images of the basis vectors, the origin stays fixed, and polygons are transformed vertex by vertex.
A linear transformation \(T\) of the plane maps each point (or position vector) \(\mathbf{x}=(x,y)\) to an image \(\mathbf{x}'=(x',y')\) by multiplying by a fixed \(2\times2\) matrix \(A\): you write \(T(\mathbf{x})=A\mathbf{x}\) and evaluate it with the row-by-column rule.
The columns of \(A\) are the images of the two basis vectors \(\mathbf{i}=(1,0)\) and \(\mathbf{j}=(0,1)\): the first column is \(T(\mathbf{i})\) and the second column is \(T(\mathbf{j})\). So if you know where \(\mathbf{i}\) and \(\mathbf{j}\) go, you can build \(A\) by placing those images as columns.
Every linear transformation is linear: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(k\mathbf{v})=k\,T(\mathbf{v})\). A consequence is that the origin maps to itself, \(T(\mathbf{0})=\mathbf{0}\); straight lines stay straight, so a translation (which moves the origin) is not a linear transformation.
To transform a polygon, apply \(A\) to each vertex and join the image points in the same order. The unit square maps to a parallelogram whose area equals \(|\det A|\), the area scale factor of the transformation.
A linear transformation applies the matrix \(A\) to the position column vector:
The columns of \(A\) are the images of the basis vectors, so \(A\) is built from where \(\mathbf{i}\) and \(\mathbf{j}\) go:
The linearity properties let you combine known images:
How to find the image of a point
- Write the point as a column vector \(\begin{pmatrix}x\\y\end{pmatrix}\), placing it to the right of the matrix \(A\).
- Multiply row by column: the top entry is \(ax+by\) and the bottom entry is \(cx+dy\).
- Simplify the arithmetic in each entry to get the image column \(\begin{pmatrix}x'\\y'\end{pmatrix}\).
- Write the image as a point \((x',y')\). For a polygon, repeat for every vertex and join the images in order; to build \(A\), place \(T(\mathbf{i})\) and \(T(\mathbf{j})\) as its columns.
Write \(A\) times the point as a column vector, then multiply row by column:
| \(\begin{pmatrix}x'\\y'\end{pmatrix}\) | \(=\) | \(\begin{pmatrix}3&1\\2&-1\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}3\times2+1\times4\\ 2\times2+(-1)\times4\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}6+4\\ 4-4\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}10\\0\end{pmatrix}\) |
The image is \((10,0)\).
The image of \(\mathbf{i}\) is the first column of \(A\) and the image of \(\mathbf{j}\) is the second column:
| \(\text{image of }\mathbf{i}=(4,-1)\) | \(\Rightarrow\) | \(\text{column }1\) |
| \(\text{image of }\mathbf{j}=(2,3)\) | \(\Rightarrow\) | \(\text{column }2\) |
| \(A\) | \(=\) | \(\begin{pmatrix}4&2\\-1&3\end{pmatrix}\) |
\(A=\begin{pmatrix}4&2\\-1&3\end{pmatrix}\).
Apply the linearity rules \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(k\mathbf{v})=k\,T(\mathbf{v})\):
| \(T(3\mathbf{u}-\mathbf{v})\) | \(=\) | \(3\,T(\mathbf{u})-T(\mathbf{v})\) |
| \(=\) | \(3\begin{pmatrix}2\\5\end{pmatrix}-\begin{pmatrix}1\\-2\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}6\\15\end{pmatrix}-\begin{pmatrix}1\\-2\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}6-1\\15-(-2)\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}5\\17\end{pmatrix}\) |
\(T(3\mathbf{u}-\mathbf{v})=(5,17)\).
The corners \((1,0)\) and \((0,1)\) are \(\mathbf{i}\) and \(\mathbf{j}\), so their images are the columns of \(A\); the origin is fixed. Multiply each corner:
| \((0,0)\) | \(\to\) | \((0,0)\) |
| \((1,0)\) | \(\to\) | \((3,0)\) |
| \((1,1)\) | \(\to\) | \((3\times1+1\times1,\ 0\times1+2\times1)=(4,2)\) |
| \((0,1)\) | \(\to\) | \((1,2)\) |
The image is the parallelogram \((0,0),(3,0),(4,2),(1,2)\). Its area is the scale factor \(|\det A|\):
| \(\det A\) | \(=\) | \(3\times2-1\times0\) |
| \(=\) | \(6\) |
The image is the parallelogram \((0,0),(3,0),(4,2),(1,2)\), with area \(6\).
Common pitfalls
Frequently asked questions
What is a linear transformation in Specialist Maths?
It is a map \(T\) of the plane that sends each point \((x,y)\) to an image using a fixed \(2\times2\) matrix \(A\), written \(T(\mathbf{x})=A\mathbf{x}\).
How do you find the image of a point under a matrix?
Write the point as a column vector to the right of \(A\) and multiply row by column: the image is \(\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}\).
Why are the columns of A the images of i and j?
Because \(A\mathbf{i}\) picks out the first column and \(A\mathbf{j}\) picks out the second column, so \(A=[\,T(\mathbf{i})\mid T(\mathbf{j})\,]\). This lets you build \(A\) from where \(\mathbf{i}\) and \(\mathbf{j}\) go.
Does the origin always stay fixed?
Yes. For any matrix \(A\), \(A\mathbf{0}=\mathbf{0}\), so a linear transformation always maps the origin to itself. A translation moves the origin and is therefore not linear.
How do you transform a polygon?
Apply the matrix to each vertex separately, then join the image points in the same order. The unit square becomes a parallelogram.
What does the determinant tell you about a transformation?
The area of an image is \(|\det A|\) times the original area, so \(|\det A|\) is the area scale factor of the transformation.