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**preview**shows pages 1-2. to view the full**8 pages of the document.**3 Linear Transformations of the Plane

Now that we’re using matrices to represent linear transformations, we’ll ﬁnd ourselves en-

countering a wide range of transformations and matrices; it can become diﬃcult to keep

track of which transformations do what. In these notes we’ll develop a tool box of basic

transformations which can easily be remembered by their geometric properties.

We’ll focus on linear transformations T:R2→R2of the plane to itself, and thus on the

2×2 matrices Acorresponding to these transformation. Perhaps the most important fact

to keep in mind as we determine the matrices corresponding to diﬀerent transformations is

that the ﬁrst and second columns of Aare given by T(e1) and T(e2), respectively, where e1

and e2are the standard unit vectors in R2.

3.1 Scaling

The ﬁrst transformation of R2that we want to consider is that of scaling every vector by

some factor k. That is, T(x) = kxfor every x∈R2. If k= 1, then Tdoes nothing. In this

case, T(e1) = e1and T(e2) = e2, so the columns of the corresponding matrix Aare e1and

e2:

A=1 0

0 1.

We call this the identity matrix (of size 2) and denote it either as I2or as Iwhen the

size is obvious. For any other scale factor kwe have

T(e1) = k

0and T(e2) = 0

k,

so the corresponding matrix is given by

A=k0

0k.

Now suppose we have two scaling maps: T1which scales by a factor of k1, and T2which

scales by a factor of k2. Then T2◦T1is the transformation that scales by a factor of k1and

then by k2, which is to say that it scales by a factor of k2k1. This means that its matrix

representation should be given by

A=k2k10

0k2k1,

and indeed we can easily check that

A2A1=k20

0k2k10

0k1=k2k10

0k2k1=A.

This agrees with our notion of matrix multiplication representing composition of linear trans-

formations.

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Figure 2: Orthogonal projection of vonto w.

3.2 Orthogonal Projection

The next linear transformation we’d like to consider is that of projecting vectors onto a line

in R2. First we have to consider what it means to project one vector onto another. Take a

look at Figure 2, where we’re projecting the vector vonto worthogonally. What we mean

by orthogonal projection is that the displacement vector projwv−vis orthogonal to the

vector w, as seen in Figure 2.

We can see that projwvwill be a scalar multiple of w, so let’s write projwv=kw. Since

we require that projwv−vbe orthogonal to w, we have

(kw−v)·w= 0.

That is,

kw·w−v·w= 0,

so

k=v·w

w·w.

This means that we have

projwv=v·w

w·ww.

Now notice that if we project vonto any vector which is a nonzero scalar multiple of w, the

resulting vector will be the same as projwv. So really we’re projecting vonto the line L

determined by w. For this reason, we write projLvfor the projection of vonto the line L.

Given a line L, we can compute projLvby ﬁrst selecting a unit vector u=hu1, u2i

through which Lpasses and then projecting vonto u. We then have

projL(e1) = proju(e1) = e1·u

u·uu=u1

1u=u2

1

u1u2

and

projL(e2) = proju(e2) = e2·u

u·uu=u2

1u=u1u2

u2

2,

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Figure 3: Rotation by θ.

so the matrix Acorresponding to the projection onto Lis

A=u2

1u1u2

u1u2u2

2.

As before, there’s a matrix product that’s worth considering here. If we apply the transfor-

mation of projecting onto Ltwice, this should be no diﬀerent than applying it once. After

the ﬁrst projection, all the vectors in R2have been mapped onto L, and projecting a vector

on Lonto Ldoes nothing. This is borne out by the fact that

A2=u2

1u1u2

u1u2u2

2 u2

1u1u2

u1u2u2

2=u2

1u1u2

u1u2u2

2=A.

We can also verify our claim that projecting a vector which already lies on Lonto Ldoes

nothing:

Aku1

ku2=ku3

1+ku1u2

2

ku2

1u2+ku3

2=ku1(u2

1+u2

2)

ku2(u2

1+u2

2)=ku1

ku2,

where we using the fact that any vector on Lhas the form hku1, ku2ifor some k.

3.3 Rotation

Next we’ll consider rotating the plane through some angle θ, as depicted in Figure 3. Because

the vector e1lies on the unit circle, so does T(e1), and T(e1) makes an angle of θwith the

x-axis. As a result, its x- and y-components are cos θand sin θ, respectively:

T(e1) = cos θ

sin θ.

At the same time, since e2makes an angle of π/2 with e1, the vectors T(e2) and T(e1)

should also have an angle of π/2 between them. So the x- and y-components of T(e2) are

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