3.4: Examples of linear transformations

A first example¶

Consider

It’s not immediately obvious what multiplying by $A$ will do. Let’s try a few inputs:

The first two outputs are the columns of $A$. The third is the sum of the two columns of $A$ – this should not be a surprise, since the act of multiplying by $A$ is a linear transformation, and so it must be true that

for all possible $\vec u$ and $\vec v$. (Here, $\vec u$ is the first column of $A$ and $\vec v$ is the second column of $A$.)

Let $\color{#3d81f6}{\vec e_1}=\begin{bmatrix}1\\0\end{bmatrix}$ and $\color{orange}{\vec e_2}=\begin{bmatrix}0\\1\end{bmatrix}$ be the standard basis vectors. To see more than individual vectors, let’s draw the square with corners $\vec0$, $\vec e_1$, $\vec e_1+\vec e_2$, and $\vec e_2$, and apply $A$ to every point in it. This is a visualization strategy we will use repeatedly in this section.

The square becomes a parallelogram! As we will see later in this section, this transformation is an example of a shear. As a function, this is the transformation

For instance, $\begin{bmatrix}2\\-1\end{bmatrix}$ becomes $\begin{bmatrix}1\\2\end{bmatrix}$.

For the next several examples, we focus on transformations from $\mathbb R^2$ to $\mathbb R^2$ – which are defined using a $2 \times 2$ matrix – because transformations in which the input and output vectors are both in $\mathbb{R}^2$ are easy to visualize. Later, we will consider transformations whose domains and codomains have different shapes.

Stretching¶

Suppose we want to stretch everything horizontally by a factor of 2, leaving the vertical coordinates alone. What should happen to $\begin{bmatrix}x\\y\end{bmatrix}$? It should become

Its associated matrix – that is, the matrix $H$ for which $T \left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = H \begin{bmatrix} x \\ y \end{bmatrix}$ – is

We could also find $H$ by following the standard basis vectors. Recall that the standard basis vectors in $\mathbb R^2$ are

At the end of Chapter 3.3, we said that if we know the outputs of a linear transformation for $\vec e_1$ and $\vec e_2$ — that is, $T(\vec e_1)$ and $T(\vec e_2)$ — we know its entire associated matrix:

Here, the first basis vector doubles in length, while the second stays where it is:

Putting these outputs into columns gives $H=\begin{bmatrix}2&0\\0&1\end{bmatrix}$, as before.

More generally, a diagonal matrix $\begin{bmatrix}a&0\\0&b\end{bmatrix}$ multiplies the first coordinate by $a$ and the second by $b$. When $a=b=2$, both coordinates double, so the whole square doubles in size. When a scale factor is negative, there is a flip as well, which we can interpret as a reflection across an axis – as we’re about to do.

Reflection¶

To reflect across the $x$-axis, keep the first coordinate and change the sign of the second:

This is slightly counterintuitive, perhaps: reflecting across the $x$-axis changes the vector’s $y$ component, not its $x$ component.

This transformation has associated matrix

For example, $\begin{bmatrix}2\\1\end{bmatrix}$ becomes $\begin{bmatrix}2\\-1\end{bmatrix}$. A point above the $x$-axis moves the same distance below it. Points already on the $x$-axis stay put.

What would change if we reflected across the $y$-axis instead? Now we would change the sign of the first coordinate, giving $\begin{bmatrix}-1&0\\0&1\end{bmatrix}$.

We’ve shown the two simplest examples of reflections – across the $x$ or $y$ axes. In a homework problem, you’ll explore how to reflect across a line that has some slope, e.g. across the line spanned by $\begin{bmatrix} 3 \\ -5 \end{bmatrix}$.

Rotation¶

What matrix rotates vectors counterclockwise by $90^\circ$ about the origin? Rather than guessing all four entries, let’s follow the standard basis vectors:

These are the columns of the transformation’s associated matrix! Therefore,

For example, $R\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}-1\\2\end{bmatrix}$.

The same idea works for any fixed angle $\theta$. A counterclockwise rotation sends

The first vector, $T(\vec e_1)$, comes from the unit circle: trigonometry tells us that the point at distance 1 from $(0,0)$ that makes an angle $\theta$ with the $x$-axis is $(\cos\theta,\sin\theta)$. The second vector, $T(\vec e_2)$, is a $90^\circ$ counterclockwise rotation from the first. Thus

where the associated matrix is

Use the slider below to change $\theta$.

Rotations preserve lengths and angles between vectors:

• Preserving length means that $\lVert R_\theta \vec x \rVert = \lVert \vec x \rVert$ for all $\vec x$. This is not true of every linear transformation: a transformation that stretches the $x$-component by 2 and keeps the $y$-component fixed doubles the length of $\vec e_1$ but leaves the length of $\vec e_2$ unchanged.

• Preserving angles means that the angle between $\vec u$ and $\vec v$ is the same as the angle between $R_\theta \vec u$ and $R_\theta \vec v$. Again, this is also not true in general of all linear transformations.

You might notice that the columns of $R_\theta$ form an orthonormal basis for $\mathbb R^2$. Their dot product is

and their lengths are

It turns out that any $2\times2$ matrix whose columns form an orthonormal basis for $\mathbb R^2$ represents a rotation, possibly combined with a reflection.

For example, consider

Its columns have dot product $\left(\frac35\right)\left(\frac45\right)+\left(-\frac45\right)\left(\frac35\right)=0$, and each has length $\sqrt{(3/5)^2+(4/5)^2}=1$. This matrix is a rotation with $\cos\theta=3/5$ and $\sin\theta=-4/5$: it rotates vectors clockwise by approximately $53.1^\circ$. Here is what it does to the unit square and standard basis vectors.

Shear¶

In this next transformation, imagine glueing one edge of the unit square to the $x$-axis, and dragging the top-right corner of it to the right or left. Such a transformation may be of the form:

Its associated matrix is

This is called a horizontal shear.

For example, $\begin{bmatrix}2\\1\end{bmatrix}$ becomes $\begin{bmatrix}3\\1\end{bmatrix}$, while $\begin{bmatrix}2\\0\end{bmatrix}$ stays unchanged. This isn’t moving everything the same distance to the right: how far a point moves depends on its second coordinate.

A stronger horizontal shear is

with associated matrix $\begin{bmatrix}1&3\\0&1\end{bmatrix}$. More generally, replacing the upper-right entry by $k$ gives the horizontal shear $T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x+ky\\y\end{bmatrix}$.

A vertical shear instead adds a multiple of the horizontal coordinate to the vertical coordinate. For example,

whose associated matrix is $\begin{bmatrix}1&0\\2&1\end{bmatrix}$.

Projection¶

Let’s return to an operation we already know: orthogonal projection. Let’s start with the simplest case, projecting onto the $x$-axis. We keep the horizontal coordinate and set the vertical coordinate to zero:

Its associated matrix is

What happens to our square this time?

In all previous examples, the unit square on the left got transformed into a parallelogram on the right. Not here: the transformation sent both $\vec e_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\vec e_2=\begin{bmatrix}0\\1\end{bmatrix}$ to vectors that live on the $x$-axis ($\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\0\end{bmatrix}$, respectively), so there is no longer a parallelogram — just a line segment.

Different inputs can have the same output: $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}1\\5\end{bmatrix}$ both become $\begin{bmatrix}1\\0\end{bmatrix}$.

We defined $P:\mathbb R^2\to\mathbb R^2$, so its codomain is $\mathbb R^2$. But the outputs it actually produces all lie on the $x$-axis – that is, its set of actual outputs is different than its set of possible outputs. There is no way for, say, $\begin{bmatrix}3\\4\end{bmatrix}$ to be the result of $P\vec x$ for any vector $\vec x\in\mathbb R^2$.

A slanted line¶

What if we project onto the line $y=x$ instead? Its unit direction vector is $\vec w=\frac1{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}$. Using the projection formula from Chapter 2,

Both output entries are the average of the two input entries, so the associated matrix is

For instance, $\begin{bmatrix}3\\1\end{bmatrix}$ projects to $\begin{bmatrix}2\\2\end{bmatrix}$. Here the original vector $\vec v$ and its transformed vector $T(\vec v)$ appear in the same plot, rather than side-by-side in separate plots as in the previous examples.

More generally, if $L=\operatorname{span}(\vec w)$ and $\vec w=\begin{bmatrix}w_1\\w_2\end{bmatrix}$ is a unit vector, then projection onto $L$ has associated matrix

For example, consider projecting onto the line spanned by $\vec w=\begin{bmatrix}3\\-5\end{bmatrix}$. This vector has length $\sqrt{34}$, so we first normalize it to get the unit direction vector

Substituting these entries into the formula gives

Thus, the transformation is

Here is the unit square before and after this projection. After the transformation, both basis vectors lie on the line spanned by $\vec w$.

A plane¶

In $\mathbb R^3$, projection onto the $xy$-plane is simple: set the third coordinate to zero.

Its associated matrix is

The matrix $Q$ has shape $3\times3$ because both the input and output vectors are in $\mathbb R^3$.

For example, $\begin{bmatrix}2\\1\\3\end{bmatrix}$ becomes $\begin{bmatrix}2\\1\\0\end{bmatrix}$. You can drag the picture below to see the perpendicular drop to the plane.

A matrix for projecting onto a plane other than the $xy$-, $xz$-, or $yz$-plane is slightly more complicated. We’ll explore these projections soon.

Lines stay lines¶

One thing you should have noticed in all of the transformations we’ve visualized is that the shape formed by transforming the standard basis vectors, e.g. the parallelogram below, always contained straight lines — never any curves.

This is a general property of linear transformations: points that lie along a line in the domain still lie along a line in the output, unless the entire line collapses to a single point.

For example, use the horizontal shear matrix $K$ again:

Follow three points on a line that does not pass through the origin as we apply $K$:

Note that all three points lived on a line before being sent through the transformation, and still live on a line after the transformation. A different line than before, yes, but still a line nonetheless.

Why does this happen? Choose a point $\vec p$ on the line and a nonzero direction vector $\vec v$. Every other point on the line is obtained by adding some multiple of $\vec v$ to $\vec p$. Because $T$ is a linear transformation, it satisfies both of the following properties:

• $T(\vec u+\vec v)=T(\vec u)+T(\vec v)$ for every pair of vectors $\vec u,\vec v$ in its domain.

• $T(c\vec v)=cT(\vec v)$ for every vector $\vec v$ in its domain and every scalar $c$.

Applying these properties gives

So every output is obtained by starting at $T(\vec p)$ and adding a multiple of the same direction vector $T(\vec v)$. The outputs still lie on a line. If $T(\vec v)=\vec0$, the entire line instead collapses to the single point $T(\vec p)$.

In many applications, keeping straight lines straight is desirable. When drawing a 3D scene on a 2D screen, for example, we want a straight building edge to appear straight in the image. We’ll explore this idea in just a moment.

Different domain and codomain¶

So far, each transformation has returned the same number of entries it received. We looked at transformations from $\mathbb R^2$ to $\mathbb R^2$, described with a $2\times2$ matrix, and transformations from $\mathbb R^3$ to $\mathbb R^3$, described with a $3\times3$ matrix. But, a transformation can have a different domain (set of possible inputs) and codomain (set of possible outputs).

For example, the function

copies the two input entries and adds their sum as a third entry. Its associated matrix is

For example,

The domain is $\mathbb R^2$, the codomain is $\mathbb R^3$, and the matrix has shape $3\times2$.

As another example, define

Its associated matrix is $B=\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}$.

This has codomain $\mathbb R^2$. Compare it with projection onto the $xy$-plane above: $\begin{bmatrix}x\\y\end{bmatrix}$ has two entries, while $\begin{bmatrix}x\\y\\0\end{bmatrix}$ has three.

Example: computer graphics¶

Computer games, and engineering software that visualize physical parts, all involve visualizing a three-dimensional world on your two-dimensional computer screen. To convert a three-dimensional world into a two-dimensional representation, we can use a linear transformation! Explore this below.

Start with the overhead view, then drag View toward Game view. Change Building height, and turn on Show basis arrows to see where the three standard basis vectors go.

In game view, the transformation uses the matrix

We choose how a one-unit step in each world direction should look on the screen.

• One step in the world $x$-direction should move 40 units right and 20 units down on the screen.

• One step in the world $y$-direction should move 40 units right and 20 units up.

• One step upward in the world $z$-direction should move 40 units up, with no horizontal movement.

These choices tell us where the three standard basis vectors go:

Put those three outputs into columns, and we get $A$. The numbers 40 and 20 are drawing choices: their ratio makes the ground directions slope by one screen unit vertically for every two units horizontally: downward for the $x$-direction and upward for the $y$-direction. The 40 in the third column sets how tall a one-unit vertical edge appears. Other choices would give a different view or scale; doubling every entry would double the size of the drawing.

For an arbitrary world point, we combine these three screen movements:

For example, the ground corner $\begin{bmatrix}1\\1\\0\end{bmatrix}$ appears at $\begin{bmatrix}80\\0\end{bmatrix}$. The roof corner directly above it has height $h$, so

The horizontal coordinate stays at 80, while the vertical coordinate is 40 times the building’s height. At height $h=1$, the roof corner appears at $\begin{bmatrix}80\\40\end{bmatrix}$.

The domain is $\mathbb R^3$ and the codomain is $\mathbb R^2$. In the overhead view, we instead use

Now the two ground directions appear horizontal and vertical, and height disappears because the third column is zero. The View slider blends this matrix with the game-view matrix; each setting gives a linear transformation. The current matrix $A$ is shown beneath the transformation in the interactive.