Linear Transformations of the Plane

The decoder ring: columns tell the whole story

A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a function satisfying two rules:

1. $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$ (preserves addition)
2. $T(c\vec{v}) = c T(\vec{v})$ (preserves scalar multiplication)

From these two rules, something spectacular follows. Pay attention — the Federation of Boring Textbook Authors buries this in chapter twelve and calls it a "theorem" instead of leading with it. Any vector $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ can be written as $x\hat{e}_1 + y\hat{e}_2$ where $\hat{e}_1 = \begin{pmatrix}1\\0\end{pmatrix}$ and $\hat{e}_2 = \begin{pmatrix}0\\1\end{pmatrix}$ are the standard basis vectors. Apply $T$ using the two rules:

$T(\vec{v}) = T(x\hat{e}_1 + y\hat{e}_2) = x T(\hat{e}_1) + y T(\hat{e}_2)$

The entire transformation is determined by where the two basis vectors land. Know $T(\hat{e}_1)$ and $T(\hat{e}_2)$, and you know everything — every damn point on the plane included. This is the decoder ring:

$\underbrace{\begin{pmatrix} T(\hat{e}_1) & T(\hat{e}_2) \end{pmatrix}}_{\text{matrix }A}\vec{v} = A\vec{v}$

Column 1 of $A$ is where $\hat{e}_1 = (1,0)^T$ goes. Column 2 of $A$ is where $\hat{e}_2 = (0,1)^T$ goes. If I hand you a $2\times 2$ matrix, you can immediately read off the images of both basis vectors from the columns, and from those two images you know the entire transformation's behavior everywhere.

What "linear" means visually

A linear transformation is characterized by what it does to the plane:

• Grid lines stay parallel and evenly spaced. Straight lines remain straight, and equally-spaced parallel lines remain equally-spaced and parallel.
• The origin stays fixed. $T(\vec{0}) = T(0 \cdot \vec{v}) = 0 \cdot T(\vec{v}) = \vec{0}$.

If either of these fails, the transformation is not linear. A rotation about a point other than the origin? Not linear. A translation (sliding everything by $(3, 2)$)? Not linear — it moves the origin. Linear algebra is exclusively the mathematics of transformations that fix the origin and preserve the grid structure.

Drag the columns of the matrix below and watch the grid deform:

The greatest hits: rotation

Here's where the Trig stratum cashes its damn check. The rotation matrix that rotates every vector counterclockwise by angle $\theta$ is:

$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

Where does this come from? Apply the decoder ring: where does $\hat{e}_1 = (1,0)^T$ go when rotated by $\theta$? It lands at $(\cos\theta, \sin\theta)^T$. Where does $\hat{e}_2 = (0,1)^T$ go? It lands at $(-\sin\theta, \cos\theta)^T$. Those are the two columns. Done.

The trig sum-formula check. If we rotate by $\alpha$ and then by $\beta$, the composition is rotation by $\alpha + \beta$. So $R_\alpha R_\beta = R_{\alpha+\beta}$. Let's multiply it out and see what we get. By the matrix product rule:

$R_\alpha R_\beta = \begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}\begin{pmatrix}\cos\beta & -\sin\beta \\ \sin\beta & \cos\beta\end{pmatrix}$

Entry $(1,1)$: $\cos\alpha\cos\beta + (-\sin\alpha)\sin\beta = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

Entry $(2,1)$: $\sin\alpha\cos\beta + \cos\alpha\sin\beta$

But $R_{\alpha+\beta}$ has entry $(1,1) = \cos(\alpha+\beta)$ and entry $(2,1) = \sin(\alpha+\beta)$.

So the product demands: $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$. These are exactly the trig angle-addition formulas. The rotation matrix doesn't just use those formulas — multiplying rotation matrices proves them, from pure linear algebra. This is what it looks like when two branches of mathematics shake hands. The trig textbook authors should be furious; we proved their formulas in a linear algebra lab.

The other transformations

Scaling by factor $a$ horizontally and $b$ vertically: $S = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ $\hat{e}_1 \to (a, 0)^T$, $\hat{e}_2 \to (0, b)^T$. Diagonal matrices are pure stretches along the coordinate axes.

Horizontal shear by factor $k$: $H = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ $\hat{e}_1 \to (1, 0)^T$ (unchanged), $\hat{e}_2 \to (k, 1)^T$ (dragged sideways). Shears preserve area — file that away for the determinant lesson.

Reflection across the $x$-axis: $F = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ $\hat{e}_1$ stays; $\hat{e}_2$ flips to $(0,-1)^T$. The column picture tells you everything.

Composition: order matters and now you can see why

Rotating by $90°$ then shearing is NOT the same as shearing then rotating. You can feel this by thinking about it geometrically — the shear drags things horizontally, and whether you've been rotated first makes a massive difference. The matrix product captures this: $R_{90°} H \ne H R_{90°}$. I cannot stress this enough. Fucking up the order here will haunt you until the eigenvalue lesson.

To apply transformation $B$ first, then $A$: compute $AB$. The rightmost matrix acts first. This is the same notation convention as function composition: $(A \circ B)(\vec{v}) = A(B(\vec{v}))$.

Degenerate transformations: when the plane gets squashed

What if both columns of the matrix point in the same direction? Then the entire plane gets collapsed onto a single line. The transformation has squashed the plane's two-dimensional structure into one dimension — a geometric felony. A vector that was not in the image has nowhere to come from — the transformation is not invertible.

This is the preview of determinants: the determinant measures how much the area scales. When the plane collapses to a line, area goes to zero, and so does the determinant. That's next lesson — and it's where this dimension-murder gets a proper name and a body count.