Linear transformations
Learn linear transformations for Year 11 Specialist Mathematics in Queensland (QCAA). 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.
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 (QCAA, Queensland) 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.