LINEAR ALGEBRA · A FIRST LOOK

Matrices that move a plane

A 2 × 2 matrix is four numbers with a geometric meaning. Tell two special vectors where to go, and every other point follows. Start with ordinary coordinates, then watch what changes.

01Two directions are enough to locate every point.

A vector is a directed displacement: (1, 0) means one step right and zero steps up. (0, 1) means zero steps right and one step up. Draw them from the origin of an ordinary Cartesian coordinate system.

THE STANDARD BASIS

I =

\begin{matrix}1\\0\end{matrix}

\begin{matrix}0\\1\end{matrix}

Column 1 → e₁Column 2 → e₂

The standard orthonormal basis

ortho: Perpendicular / orthogonal: the angle between them is 90°.
normal: Normalized: each vector has magnitude (length) 1.
basis: Two independent directions. Scaling and adding them reaches every point, with one unique pair of coefficients.

\begin{pmatrix}x\\y\end{pmatrix}=x\underbrace{\begin{pmatrix}1\\0\end{pmatrix}}_{\mathbf e_1}+y\underbrace{\begin{pmatrix}0\\1\end{pmatrix}}_{\mathbf e_2}

Read down each column of I: the first is e₁, the second is e₂. This is the identity matrix. As a transformation it sends each basis vector back to itself, so every point stays where it was.

I\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\y\end{pmatrix}

02Change the columns. Move the whole plane.

In the starting example, the first column tells e₁ to go to u = (2, 1); its second tells e₂ to go to v = (1, 2). The dashed arrows mark those destinations before anything moves. Press play, then watch the two families of grid lines follow their basis vectors.

FIRST, LOCATE THE TWO DESTINATIONS0%

Animation progress0%

Moving point

P'=\begin{pmatrix}1\\1\end{pmatrix}

Shaded area 1

The destination matrix

A =

u=\begin{pmatrix}2\\1\end{pmatrix}

v=\begin{pmatrix}1\\2\end{pmatrix}

u and v are independent, so they form a new basis. But they are not perpendicular and neither has length 1: a basis need not be orthonormal.

Follow one point

x and y count how much of each basis vector to add. Try a negative value: that step goes in the opposite direction.

Original x · first vector1

Original y · second vector1

A\begin{pmatrix}1\\1\end{pmatrix}=1\,\begin{pmatrix}2\\1\end{pmatrix}+1\,\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}3\\3\end{pmatrix}

This equation gives the final point. The solid coloured step followed by the dashed coloured step shows the same addition on the moving grid.

What moved, and what did not?

The faint Cartesian axes are our fixed ruler. We actively move the points and the coloured grid; we are not moving that ruler. In the example, P = (1, 1) becomes P′ = (3, 3) on the fixed axes. Relative to the new basis, the image still means “one u plus one v”. Changing the coordinates of an unmoved point would be a different operation.

A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\qquad A\begin{pmatrix}x\\y\end{pmatrix}=x\begin{pmatrix}a\\c\end{pmatrix}+y\begin{pmatrix}b\\d\end{pmatrix}

This is why two columns determine the whole transformation: every vector is a scaled sum of the original basis vectors. A linear transformation preserves those sums and scale factors. Straight lines remain straight (or collapse), parallel grid lines remain parallel or coincide/collapse, and the origin stays put.

Four transformations, one idea.

The coloured arrows are still the columns of the matrix. The shaded unit square reveals area; the circle reveals stretching; the letter F makes mirror reversal visible. Each animation starts with the original shapes. Use its progress slider to compare any moment.

Rotation

Turn every point around the origin by the same angle. A circle stays a circle, and the two basis vectors remain perpendicular with length 1.

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R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}

Target angle θ · positive = counterclockwise90°

Area now 1 ×Angle now0°

Length, angle and area are preserved throughout. Negative angles rotate clockwise.

Shear

Imagine sliding the layers of a deck of cards. Each horizontal row moves by k times its height y. Height stays fixed; squares lean into parallelograms.

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S(k)=\begin{pmatrix}1&k\\0&1\end{pmatrix}\quad\begin{cases}x'=x+ky\\y'=y\end{cases}

Target shear factor k1.5

Area now 1 ×Final area1 ×

Base and perpendicular height stay the same, so area stays 1. Angles and most lengths change.

Reflection

Use the vertical y-axis as a mirror. Right becomes left, while height stays the same. Watch the letter F: its handedness reverses.

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F=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\quad\begin{cases}x'=-x\\y'=y\end{cases}

The endpoint is a reflection. To animate the change, this demo shrinks the horizontal scale through 0 and expands it on the other side. The flattened middle frame is not itself a reflection or a rotation.

Area now 1 ×Final area1 ×

At the end, lengths and areas are unchanged, but orientation is reversed.

Stretch & compress

Multiply horizontal and vertical distances separately. Unequal factors turn a circle into an ellipse. A factor below 1 compresses that direction.

Animation progress0%

D=\begin{pmatrix}s_x&0\\0&s_y\end{pmatrix}\quad\begin{cases}x'=s_xx\\y'=s_yy\end{cases}

Horizontal factor sₓ2

Vertical factor sᵧ0.65

Area now 1 ×Final area1.3 ×

Area is multiplied by sₓ × sᵧ. Doubling both directions multiplies area by 4, not 2.