Topic 4.13 Notes – Matrices as Functions

What a Matrix Transformation Is

A $2\times2$ matrix can act on a vector in $\mathbb R^2$. If

$$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} \quad\text{and}\quad \mathbf v=\begin{bmatrix}x\\y\end{bmatrix},$$

then the transformation is

$$L(\mathbf v)=A\mathbf v$$

which means

$$(x,y)\mapsto (a_{11}x+a_{12}y,\;a_{21}x+a_{22}y).$$

So these are all the same transformation:

• the matrix $A$
• the rule $(x,y)\mapsto(\dots,\dots)$
• a verbal description like “stretch, reflect, or rotate vectors”

A quick example:

• $L(x,y)=(2x-3y,\;x+4y)$
• matrix form is $\begin{bmatrix}2&-3\\1&4\end{bmatrix}$

The input vector is the preimage. The output is the image.

A linear transformation has no constant term. That means the origin stays fixed:

• linear: $(x,y)\mapsto(2x-y,\;x+3y)$
• not linear: $(x,y)\mapsto(2x-y+4,\;x+3y)$

That $+4$ shifts points, so it is outside this AP topic.

Unit Vectors, Columns, and Key Geometric Meaning

The whole matrix is determined by what it does to the standard unit vectors

$$\mathbf e_1=\begin{bmatrix}1\\0\end{bmatrix}, \qquad \mathbf e_2=\begin{bmatrix}0\\1\end{bmatrix}.$$

Every vector can be written as $x\mathbf e_1+y\mathbf e_2$, so

$$L(x,y)=xL(\mathbf e_1)+yL(\mathbf e_2).$$

That is why:

• first column $=L(\mathbf e_1)$
• second column $=L(\mathbf e_2)$

If $L(\mathbf e_1)=\begin{bmatrix}-1\\2\end{bmatrix}$ and $L(\mathbf e_2)=\begin{bmatrix}3\\1\end{bmatrix}$, then

$$A=\begin{bmatrix}-1&3\\2&1\end{bmatrix}.$$

Common mistake: putting those images in rows. They go in columns.

In the diagram, those two column vectors create the image of the unit square as a parallelogram.

That picture leads straight to rotations and determinants.

Rotation matrices

A counterclockwise rotation by angle $\theta$ uses

$$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}.$$

Its unit-vector images are

• $\mathbf e_1\mapsto(\cos\theta,\sin\theta)$
• $\mathbf e_2\mapsto(-\sin\theta,\cos\theta)$

The rotation diagram tracks exactly those two images of the basis vectors.

So any vector follows the rule

$$(x,y)\mapsto(x\cos\theta-y\sin\theta,\;x\sin\theta+y\cos\theta).$$

Clockwise by $\theta$ is counterclockwise by $-\theta$. On tests, students often miss the negative sign in the upper-right entry.

Determinant as Area Change and Invertibility

For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$,

$$\det(A)=ad-bc.$$

The absolute value tells you the area scale factor:

• $|\det(A)|>1$ area increases
• $0<|\det(A)|<1$ area decreases
• $|\det(A)|=1$ area stays the same
• $\det(A)=0$ area collapses to $0$
• $\det(A)<0$ orientation reverses

This connects to the unit-square picture because the image parallelogram has area $|\det(A)|$.

One trap here: area preserved does not mean length preserved. A matrix with determinant $1$ is not automatically a rotation.

Composition and Inverse Transformations

If $F(\mathbf v)=A\mathbf v$ and $G(\mathbf v)=B\mathbf v$, then

$$(G\circ F)(\mathbf v)=(BA)\mathbf v.$$

The matrix on the right happens first. That is the order issue students lose points on most often.

To build a composition from words:

1. Identify which transformation happens first.
2. Write each matrix.
3. Multiply in composition order.
4. Use the product on the vector if needed.

Since matrix multiplication is usually not commutative, $AB\ne BA$.

The identity transformation leaves every vector alone:

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$

An inverse transformation undoes a matrix transformation. If $L(\mathbf v)=A\mathbf v$, then

• $L^{-1}(\mathbf v)=A^{-1}\mathbf v$ when $A$ is invertible
• invertible exactly when $\det(A)\ne 0$

A rotation’s inverse is rotation by $-\theta$. If $\det(A)=0$, information is lost, so no inverse exists.