LINEAR ALGEBRA · A FIRST LOOK

Matrices that move a plane

When a picture on a screen is rotated, stretched or slanted, every point of the picture moves at once. A rule that takes in a point and gives back another point is like a function. On the function page one input gives one output; here the input and output are points instead of single numbers. If we had to record where every single point goes, the description would never end. Fortunately, for transformations that keep the grid lines straight, parallel and evenly spaced, with the origin fixed, two arrows are enough. Such arrows are called vectors, and they carry the whole picture.

Keep the point $(2,1)$ in view: two steps right and one step up take us there from the origin. Each step uses one of two basic arrows. Now move those arrows, but still take two copies of the first and one of the second. That tells us where this point goes. You can keep following it and check that the same combination is being carried along. The two columns of a matrix simply record the new basic arrows. With those columns attached to something we can see, the whole grid becomes easier to follow without memorising where every point goes.

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 right and one up. Draw both arrows from the origin. We will record each arrow by putting its horizontal coordinate above its vertical coordinate. Placing these two coordinate columns side by side makes a rectangular array of numbers called a matrix. A row runs across, and a column runs down. The colours below keep each column attached to its arrow.

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 $e_1$ , the second is $e_2$ . 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 sends $e_1$ to $u=(2,1)$ and the second sends $e_2$ to $v=(1,2)$ . We define the new position of every other point by keeping its original amounts of these two arrows. For example, one $e_1$ plus one $e_2$ becomes one $u$ plus one $v$ . The dashed arrows mark the destinations before anything moves. Press play and follow the same addition as the arrows change.

FIRST, LOCATE THE TWO DESTINATIONS0%

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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.

Let us build the matrix from its two columns. A unit vector is an arrow of length 1. On the unit circle, $\cos\theta$ is the horizontal coordinate and $\sin\theta$ is the vertical coordinate at angle $\theta$ from the positive horizontal axis. The first basis arrow $(1,0)$ therefore moves to $(\cos\theta,\sin\theta)$ .

The second basis arrow starts a quarter-turn ahead of the first. It must stay a quarter-turn ahead after we rotate both by the same angle. A counterclockwise quarter-turn sends a rightward displacement to an upward one, and an upward displacement to a leftward one. Applied to $(\cos\theta,\sin\theta)$ , this gives $(-\sin\theta,\cos\theta)$ . These two destinations become the matrix columns.

\begin{gathered}\begin{pmatrix}x'\\y'\end{pmatrix}\\=x\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}+y\begin{pmatrix}-\sin\theta\\\cos\theta\end{pmatrix}\\=\begin{pmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{pmatrix}\end{gathered}

We can check the length claim rather than relying on the picture. Write $c=\cos\theta$ and $s=\sin\theta$ to shorten the expansion. The unit-circle radius is 1, so Pythagoras gives $c^2+s^2=1$ . Squaring the two new coordinates produces opposite cross terms, which cancel.

\begin{aligned}&(cx-sy)^2+(sx+cy)^2\\&=c^2x^2-2csxy+s^2y^2\\&\quad+s^2x^2+2csxy+c^2y^2\\&=(c^2+s^2)x^2+(c^2+s^2)y^2\\&=x^2+y^2\end{aligned}

The squared length is unchanged; its nonnegative square root is therefore unchanged. Now take two arrows $u=(a,b)$ and $v=(d,e)$ . We also want to check their angle. Their dot product is the sum of matching coordinate products. Expand it after the turn: the two mixed pairs cancel.

\begin{aligned}&(ca-sb)(cd-se)\\&\quad+(sa+cb)(sd+ce)\\&=c^2ad-csae-csbd+s^2be\\&\quad+s^2ad+scae+scbd+c^2be\\&=(c^2+s^2)ad+(c^2+s^2)be\\&=ad+be\end{aligned}

We have shown that turning our coordinate directions changes neither this product nor the lengths. Choose the horizontal direction along $u$ . If $U$ and $V$ are the two lengths and $\varphi$ is the angle between them, the coordinates are $u=(U,0)$ and $v=(V\cos\varphi,V\sin\varphi)$ , by the unit-circle definitions. Their dot product is $U\times V\cos\varphi +0\times V\sin\varphi=UV\cos\varphi$ . Dividing by UV is allowed when both arrows are nonzero.

For nonzero arrows the dot product divided by the product of their lengths is the cosine of their angle: it measures the part of one unit arrow along the other. Both the numerator and denominator stay the same, so this cosine, and hence the angle between 0 and π, stays the same. A parallelogram keeps its base length and its height, which is its other side length times the sine of this angle. Its area, base times height, is unchanged as well.

Shear

Start with the grid row at y=1. With k=1 it moves 1 unit right; the row at y=2 moves 2 units right. Points on the horizontal axis have y=0 and stay fixed. In general, the horizontal coordinate changes from x to x+ky, while the vertical coordinate stays y. This change is called a shear.

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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.

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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.

Matrices that move a plane

01Two directions are enough to locate every point.

The standard orthonormal basis

02Change the columns. Move the whole plane.

The destination matrix

Follow one point

What moved, and what did not?

Four transformations, one idea.

Rotation

Shear

Reflection

Stretch & compress

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