# Linear Transformations, and thier properties

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Professor 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:RnRmbeafunction that takesin
avector xand outputs the result of multiplying it bymatrix A:
fA(x)=Ax for all xRn
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, yRn(indeed: A(x+y)=Ax +Ay).
Similarly every suchfunction satisﬁes fA(λx)=λfA(x)for all λRand
xRn.
It turns out that every function that satisﬁes these twoproperties is of
the form fAfor some matrix A:
Theorem. Suppose f:RnRmsatisﬁes
f(x+y)=f(x)+f(y)for allx, yRn
f(λx)=λf(x)for allλR,xRn
Then thereexists amatrix Asuch that f(x)=Ax for allxRn.
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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Unlock all 6 pages and 3 million more documents. 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 30and thus f(x
y)=cos 30sin 30
sin 30cos 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:VWsatisfying
f(v1+v2)=f(v1)+f(v2)for allv1,v2V
f(λv)=λf (v)for allλK,vV
There are plentyof importantexamples of lineartransformations:
1. Multiplication bymatrix
f:KnKm,f(v)=Av for some ﬁxed m×nmatrix Awith coeﬃ-
cients in K
2
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