Introduction

Linear transformations are one of the most powerful ideas in linear algebra.
You already know how to compute a matrix–vector product.
Now we take the next step: understanding matrices as functions that *act* on space.

Key ideas we’ll explore:

• A matrix is not just an array of numbers—it does something to vectors.
• Linear transformations stretch, rotate, reflect, shear, or project space.
• Every linear transformation in $\mathbb{R}^n$ can be represented by a matrix (once a basis is chosen).

What Is a Linear Transformation?

A linear transformation is a function $T$ between vector spaces that satisfies:

• Additivity: $$T(u + v) = T(u) + T(v)$$
• Homogeneity: $$T(cu) = c\,T(u)$$

If both hold, the function is linear.

Examples of linear transformations:

• Scaling: $T(x) = 3x$
• Rotation in the plane
• Reflection across a line
• Projection onto an axis

Non‑examples:

• $T(x) = x + (1,0)$ (fails additivity)
• $T(x) = x^2$ (fails homogeneity)

Matrices as Linear Transformations

You already know how to compute $A x$.
Now we reinterpret it:

• Every matrix $A$ defines a linear transformation $T_A(x) = A x$.

Why this matters:

• It lets us visualize what a matrix does.
• It connects algebra (matrices) with geometry (transformations).
• It helps us understand deeper concepts like eigenvectors, rank, and dimension.

Geometric Intuition

A matrix can:

• Stretch space
  e.g., $\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}$ doubles horizontal length.
• Compress space
  e.g., $\begin{pmatrix}1/2 & 0 \\ 0 & 1/2\end{pmatrix}$ shrinks everything.
• Shear
  e.g., $\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$ slants shapes sideways.
• Rotate
  e.g., $$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
• Reflect
  e.g., across the $x$‑axis: $$\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$$

Don't worry about understanding how these different transformations work. All you need to understand is that each matrix has a “signature” effect on space.

How Matrices Act on Basis Vectors

A powerful viewpoint:

• A matrix is completely determined by what it does to the basis vectors.

For $\mathbb{R}^2$:

• $A e_1$ is the first column of $A$
• $A e_2$ is the second column of $A$

So the columns of $A$ tell you:

• Where the $x$‑axis goes
• Where the $y$‑axis goes

This makes visualization much easier.

Composition of Linear Transformations

If $T$ and $S$ are linear transformations, then:

• Applying $T$ then $S$ corresponds to matrix multiplication.
• That is: $$S(T(x)) = (S T) x$$

This is why matrix multiplication is defined the way it is.

Common Examples

1. Scaling

$$A = \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}$$ Effect: stretches space by factor 3.

2. Projection onto the $x$‑axis

$$P = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$$ Effect: collapses all points onto the $x$‑axis.

3. Reflection across the line $y = x$

$$R = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$$

Exercises

1. Compute the linear transformation $A x$ for
   $A = \begin{pmatrix}2 & 1 \\ 0 & 3\end{pmatrix}$ and $x = (1,4)$.

   Solution

   $$A x = (2\cdot1 + 1\cdot4,\; 0\cdot1 + 3\cdot4) = (6,12)$$
2. Describe in words the geometric effect of
   $A = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}$.

   Solution

   
   The matrix shears the plane horizontally: it keeps $y$ the same but shifts $x$ by $2y$.
3. Determine whether the transformation
   $T(x,y) = (2x + y,\, 3y + 1)$
   is linear.

   Solution

   
   Not linear: the constant $+1$ breaks homogeneity.
4. Compute the image of the basis vectors $e_1$ and $e_2$ under
   $A = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$.

   Solution

   • $A e_1 = (0,1)$
   • $A e_2 = (-1,0)$

   This is a $90^\circ$ rotation.
5. For the matrix
   $A = \begin{pmatrix}3 & 0 \\ 0 & -2\end{pmatrix}$,
   describe the combined geometric effect.

   Solution

   
   The matrix stretches by 3 in the $x$‑direction and reflects + stretches by 2 in the $y$‑direction.
6. True or false: The function $T(x,y) = (x+y,\, xy)$ is linear.

   Solution

   
   False: the $xy$ term is nonlinear.
7. Compute the composition $B(Ax)$ for
   $A = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$,
   $B = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$,
   and $x = (2,3)$.

   Solution

   
   First compute $A x = (2+3,\; 3) = (5,3)$.
   Then $B(Ax) = (10,6)$.