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Tuesday18/01/2011, Lecture notes byY. Burda

1Linear Transformations

In the previous twolectures wehavefound the abstractions of notions of

numbers and vectors in Rn.The abstractions were elements of aﬁeld and

vectors in an abstract vector space respectively.In this lecture weare going

to ﬁnd the rightabstraction of the notion of amatrix.

When weare solving asystem of linear equations Ax =bwearen’t

interested in the matrix Aitself, weare only interested in what is the result

of multiplying this matrix byavector x.Let’s formalize this situation alittle

bit:

Given an m×nmatrix Alet fA:Rn→Rmbeafunction that takesin

avector xand outputs the result of multiplying it bymatrix A:

fA(x)=Ax for all x∈Rn

For instance if A=(1 2

3 4)then f(x

y)=(1 2

3 4) (x

y)=x+2y

3x+4y.

Of course not everyfunction from Rnto Rmis of suchform. For instance

every function whichis of theform fAfor some matrix Asatisﬁes fA(x+y)=

fA(x)+fA(y)for all vectors x, y∈Rn(indeed: A(x+y)=Ax +Ay).

Similarly every suchfunction satisﬁes fA(λx)=λfA(x)for all λ∈Rand

x∈Rn.

It turns out that every function that satisﬁes these twoproperties is of

the form fAfor some matrix A:

Theorem. Suppose f:Rn→Rmsatisﬁes

•f(x+y)=f(x)+f(y)for allx, y∈Rn

•f(λx)=λf(x)for allλ∈R,x∈Rn

Then thereexists amatrix Asuch that f(x)=Ax for allx∈Rn.

Moreover, the matrix Ais the matrix whose columns aref(e1),...,f(en),

wheree1,...,enarethe standardbasis vectors for Rn:e1=1

.

.

.

0,...,en=

0

.

.

.

1

1

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Proof. Let x=x1

.

.

.

xn=x1e1+. . . xnen.

Then f(x)=f(x1e1+. . . +xnen)=x1f(e1)+. . . +xnf(en)(here weused

the twoproperties that fsatisﬁes). If f(e1)=a1,1

.

.

.

am,1,...,f(en)=an,1

.

.

.

an,m ,

then f(x)=x1a1,1

.

.

.

am,1+. . . +xnan,1

.

.

.

an,m =a1,1... an,1

.

.

.....

.

.

a1,m ... an,m x1

.

.

.

xn,i.e. f(x)=

Ax for the matrix A=a1,1... a1,n

.

.

.....

.

.

am,1... am,n whose columns are f(e1),...,f(en).

So instead of speaking of m×nmatrices, wecan speak of functions from

Rnto Rmsatisfying f(x+y)=f(x)+f(y)and f(λx)=λf(x).

For instance let fbethe function from R2to R2that rotates anypoint

in theplane around the origin by30 degrees counterclockwise. The condition

f(x+y)=f(x)+f(y)can beveriﬁed geometrically: fpreserves shapes of

parallelograms and parallelogramsis what weuse to add twovectors in the

plane. Similarly the second condition is very easy to verify: the result of

scaling avector and then rotating it is the same as the result of ﬁrst rotating

it and then scaling.

Thus the rotation function fis given bymultiplication byamatrix whose

columns are f(1

0)andf(0

1). Using some simple geometry wecan ﬁnd that

f(1

0)=(cos 30◦

sin 30◦)and f(0

1)=−sin 30◦

cos 30◦and thus f(x

y)=cos 30◦−sin 30◦

sin 30◦cos 30◦(x

y)for

anypoint (x

y)in the plane.

This description of matrices as functions satisfying some conditions gen-

eralizes easily to abstract vector spaces:

Deﬁnition. Alinear transformation ffrom vector spaceVover aﬁeld K

to avector spaceWover the same ﬁeld is afunction f:V→Wsatisfying

f(v1+v2)=f(v1)+f(v2)for allv1,v2∈V

f(λv)=λf (v)for allλ∈K,v∈V

There are plentyof importantexamples of lineartransformations:

1. Multiplication bymatrix

f:Kn→Km,f(v)=Av for some ﬁxed m×nmatrix Awith coeﬃ-

cients in K

2

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