This

**preview**shows pages 1-2. to view the full**8 pages of the document.**Lecture 7notes byY. Burda

1Isomorphism of vector spaces

Wefeel that the vector space of quadratic polynomialsis “the same” as the

vector space of triples of their coeﬃcients: instead of apolinomial a+bx+cx2

wecan record the coeﬃcients a

b

c;if wewantto add twopolynomials, we

can instead add the triples oftheir coeﬃcientandif wewantto multiply

apolynomial byanumber, wecan instead multiply its coeﬃcients bythis

number. As far as vector space structure goes, this isenough —in an abstract

vector space all wecan do is add vectors and multiply them byscalars. The

technical word for having vector spaces, whichare “the same” is having

“isomorphic vector spaces”.

Let’s try to formulate precisely when twovector spaces Vand Ware

isomorphic: wewantto associate to any vector of Vavector of Win such

awaythat the vector associated to v1+v2will bethe sum of the vectors

associated to v1and v2.Wealso wantthat the vector associated to λ·v

will bethe product of λand the vector associated to v.Moreover, wewant

exactly one vector of Vto beassociated to any vector in W.

Thus wearriveat the following deﬁnition:

Deﬁnition. Vectors spaces Vand Wover the same ﬁeld Kare callediso-

morphic if thereexists an invertible linear transformation T:V→W.Such

atransformation is calledan isomorphism of Vand W.

Example: Pn(K)is isomorphic to Kn+1 with the isomorphism T(a0+

a1x+. . . +anxn)=a0

.

.

.

an.

Wecan generalize this example and provethe following theorem:

Theorem. If dim V=n,then Vis isomorphic to Kn.

Proof. Choose abasis A=(v1,...,vn)of Vand let T:V→Knbethe

transformation that sends eachvector to its coordinates relativeto A:T(v)=

[v]A.This transformation is linear (exercise) andis invertible: its inverse is

T−1λ1

.

.

.

λn=λ1v1+. . . +λnvn.

1

www.notesolution.com

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

Example: construct an isomorphism of V={x

y

z∈=C|x+y+z=0}

with C2.Solution: let =(1

0

−1,0

1

−1)beabasis of V.Then the trans-

formation Tdescribed in the proof of the theorem abovesends the vec-

tor x

y

z=x1

0

−1+y0

1

−1to its coordinates x, yrelativeto the basis A:

Tx

y

z=(x

y).

It mightseem that T“forgets” the zcoordinate and thus isn’t invertible,

but it isn’t so: knowing the xand y-coordinates of avectorin V,its z-

coordinate can bereconstructed: z=−x−y.Thus the inverse of Tis

T−1(x

y)=x

y

−x−y.

Wecan generalize the statementof the previous theorem slightly:

Theorem. Vector spaces Vand Wover the same ﬁeld Kareisomorphic if

and only if dim V=dim W

Proof. Wecan use the ﬁrst theorem weproved: if T:V→Knis an iso-

morphism (n=dim V=dim W)and S:W→Knis an isomorphism, then

S−1◦T:V→Wis an isomorphism from Vto W.

Weprefer however to givean explicit construction of this isomorphism:

if (v1,...,vn)is abasis of Vand (w1,...,wn)is abasis of W,then the map

R:V→W,R(λ1v1+. . .+λnvn)=λ1w1+. . .+λnwnis an isomorphism.

Example: ﬁnd an isomorphism of V={t

s

r∈R2|t+2s+3r=0}with

W={a+bx +bx2|a, b∈R

Solution: let A=(−2

1

0,−3

0

1)beabasis for Vand let B=(1,x+x2)

beabasis of W.Let T:V→Wbethe transformation that sends the

vector t

s

r∈Vwith coordinates s, rrelativeto the basis Ato the vector

s·1+r·(x+x2)∈Whaving thesame coordinates, but relativeto B:

Tt

s

r=s+rx+rx2.This transformation is an isomorphism, with the inverse

transformation sending s+rx+rx2backto s·−2

1

0+r·−3

0

1=−2s−3r

s

r∈V.

2Linear operators

We call alinear transformation from avector space to itself alinear operator:

T:V→V.

2

www.notesolution.com

###### You're Reading a Preview

Unlock to view full version

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

If T,S:V→Vare linear operators, then wecan form linear operators

T+S,λT,TS,ST,T8,2I+(TS)5(where Istands for the identityoperator:

I(v)=vfor anyv∈V).

Given abasis Aof Vwesaythat the n×nsquare matrix [T]A,Ais the

matrix of Trelativeto the basis A(note that we choose the same basis in

the source and in the target vector space V).

If we choose adiﬀerentbasis Bfor V,then

[T]B,B=[I]B,A[T]A,A[I]A,B

i.e. [T]B,B=P−1[T]A,AP.

Wecall twomatrices with this propertysimilar or conjugate:

Deﬁnition. Two n×nmatrices Aand Bare calledsimilar if thereexists

an invertible matrix Psuch that B=P−1AP .

Weshowed abovethat twomatrices are similar if and only if they are the

matrices of the same operator relativeto twodiﬀerentbases.

One this wecan do with operators, whichwecan’t do with linear trans-

formations in general is compute their determinants:

Deﬁnition.

det T=det [T]A,A

whereAis any basis of V.

It seems that the number weget depends onthe choice of the basis A,but

in fact it doesn’t: ifBis adiﬀerentbasis, then det [T]B,B=det([I]B,A[T]A,A[I]A,B).

Wecanuse the fact that the determinantis multiplicativeto write this as

det [I]B,Adet [T]A,Adet [I]A,B=det [T]A,A,as det [I]A,B=1/det [I]B,A.Thus

wehaveproved that det [T]B,B=det [T]A,A.

3Eigenvalues and eigenvectors

Westart with ametaphor that should explain whyeigenvalues and eigen-

vectors are useful. Suppose that weare observing amagician that keeps

performing thefollowing trick: in apile of wooden and steel balls he dou-

bles the number of wooden balls. It is extremely easy to predict howmany

wooden and steel balls there are going tobein the pile after 5such perfor-

mances, if weknowhowmanythere were initially —the number of steel

3

www.notesolution.com

###### You're Reading a Preview

Unlock to view full version