Worksheet 13 (March 3)

DIS 119/120 GSI Xiaohan Yan

1 Review

DEFINITIONS

• coordinate mapping;

• matrix of linear transformation relative to bases on domain and codomain;

METHODS AND IDEAS

Theorem 1. Any two vector spaces of the same dimension are isomorphic,

since any vector space of dimension n is isomorphic to Rnunder the coordinate

mapping under a basis.

Note that vector spaces of di↵erent dimensions can never be isomorphic. This

is because isomorphism preserves linear independence and spanning property,

and thus always sends a basis to a basis. But bases of vector spaces of di↵erent

dimensions contain di↵erent number of vectors.

2 Problems

Example 1. True or false. In the last three statements, S denotes the vector

space of all smooth (infinitely di↵erentiable) functions over r0, 1s. In other

words, you do not need to worry about di↵erentiability of elements of S

( ) There exists a basis B of P2 such that 1`x has coordinate p1, 1, 1qT while

x ` x2has coordinates p´3,´3,´3q.

( ) Let V be a 5-dimensional vector space and v1,v2 be two linearly inde-

pendent vectors in V , then there exists three other vectors u1,u2,u3 in Vsuch that tv1,v2,u1,u2,u3u is a basis of V .

( ) LetW be a 2-dimensional vector space and v1,v2,v3,v4,v5 be five vectors

in W , then we can take two of these vectors to form a basis of W .

1

Recall T.IR 11pm

basisBofV bi ibis Thi't_AP

V Bpi isomorphismA Tei Tien's

cbittabi al e

r aiKil

T v w B Sbi a III T V W TB E e ki aiy BR a ke llwtc p.TTelv7B1RnslRmlv7BgnenToymatixlTiiYe BITIC bide i Fib'Re

Evibituabi't vInMinThi's Tiv.ioi rnbi

uTibTii vnTlbnT

ITNYKr.fiDci tvnTTlbi'DoLus Hibiya Embiid

Rns

BT p B 1123isoT r FIB F T

I memorizethestatement

i i lthisstatementuouidbetrueifspar.suivIgW3Tiu7tTiwy F

anyfdvectorIdea wecanjustcheckthisfoyps.dspauisiso.to113in Ingeneral theEuclidean

354in Eep Wecanalwayscomplete T twot.I.coumnvectorsotthesamed.m.asetofL2vectorsinV xp columnvectorsspanks

3 4 1 2 04010 toabasis u 7.5pivotcolumnsimpossible wecanalways removeredundancy wecandeterminepivotcolumnsbytowreductions

fromasetofspanningvectossofu viiiwillbothbepivotaltoforma basis

( ) Let A,B be two bases of the vector space V , then for any vector v P V ,

rvsA “ rBsA ¨ rvsB,

where rBsA is the square matrix whose columns are A-coordinate vectors

of the vectors of B.

( ) There is no isomorphism T : P2 Ñ R2.

( ) Let T : S Ñ S be the linear transformation T pfpxqq “ f 1pxq, then T is

surjective.

( ) Let T : S Ñ S be the linear transformation T pfpxqq “ ≥x0 fpsqds, then T

is surjective.

( ) Let T : S Ñ S be the linear transformation T pfpxqq “ f2pxq, then T has

a two-dimensional kernel.

Example 2. Let E “ te1, e2u and B “ tb1.b2u be two di↵erent bases of R2

where

e1 “ˆ1

0

˙, e2 “

ˆ0

1

˙,b1 “

ˆ2

5

˙,b2 “

ˆ1

3

˙.

(a) Let x “ˆ´1

´1

˙. Write x as a linear combination of b1 and b2.

(b) Compute rxsE and rxsB.(c) Let B “

ˆ2 1

5 3

˙. Check that rxsE “ ArxsB. Can you explain the reason

behind this?

(d) Now let’s generalize this result. Let A “ ta1.a2u be yet another basis of R2.

Given that

rb1sA “ˆ23

45

˙, rb2sA “

ˆ89

67

˙,

find rxsA.

2

B dT V W I 1ha a B IMBm

n

B a f billa Tibi'Da

SpecialcaseTeid V Da III B IMB

Tlbia11 1 1033

coordinatechangelbasechange

ITI a MIA

BB

car I'ecibitabi lb FIB d B A CHIA ARmt

Afd BtwI t IH t's Hk t 1 Ff

basesofthesame Eb H IIe B IMB Ma IBIDIIIBvectorspacerc z c 3

then B lsoxIzbTt3 tiles I 143811 34

III d Bethree E 5113 II HetfieldHBbasesthen

11M 51 HB 30153

beats

Example 3. One more question about base change. Let A “ ta1,a2,a3u and

B “ tb1.b2,b3u be two di↵erent bases of R3, where

a1 Ҭ

˝1

0

0

˛

‚,a2 “¨

˝1

1

0

˛

‚,a3 “¨

˝1

1

1

˛

‚,b1 “¨

˝1

1

1

˛

‚,b2 “¨

˝0

1

1

˛

‚b3 “¨

˝0

0

1

˛

‚.

(a) Find raisB for i “ 1, 2, 3.(b) If rxsA = p1, 1, 1qT , find rxsB.(c) If rysA “ rysB, find y.

Example 4. Consider the linear transformation T : P2 Ñ M2ˆ2pRq defined as

T pa ` bx ` cx2q “ˆ3a ` b b ` c´2b a ` b ` c

˙.

(a) Consider the basis B “ t1`x, x, x2´xu of P2 and the basis E “ tE11, E12, E21, E22uof M2ˆ2pRq. Compute BrT sE .(b) Consider instead the basis C “ tC1, C2, C3, C4u of M2ˆ2pRq, where

C1 “ˆ0 0

0 1

˙, C2 “

ˆ4 1

´2 2

˙, C3 “

ˆ1 1

´2 1

˙, C4 “

ˆ´1 0

2 0

˙.

Compute BrT sC .(c) Is T one-to-one? Onto?

3