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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 R n under 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 P 2 such that 1 ` x has coordinate p1, 1, 1q T while x ` x 2 has coordinates p´3, ´3, ´3q. ( ) Let V be a 5-dimensional vector space and v 1 , v 2 be two linearly inde- pendent vectors in V , then there exists three other vectors u 1 , u 2 , u 3 in V such that tv 1 , v 2 , u 1 , u 2 , u 3 u is a basis of V . ( ) Let W be a 2-dimensional vector space and v 1 , v 2 , v 3 , v 4 , v 5 be five vectors in W , then we can take two of these vectors to form a basis of W . 1 Recall T.IR 11pm basis B of V bi ibis Thi't _A P V Bpi isomorphism A Tei Tien's c bit tabi al e r ai Kil T vw B Sbi a III T V W T BE e ki aiy BR a ke llwtc p.TT e lv7B1Rn slRmlv7BgnenToymatixlTiiYe BITIC bide i Fib'Re E vibituabi't v In Min Thi's Tiv.io i rnb i uTibTii vnTlbnT ITNYK r.fi Dc i tvnTTlbi'Do Lus Hibiya Embiid Rns B T p B 1123 iso T r FIB F T I memorize the statement i i l this statement uouidbetrueifspar.su ivIg W3 Tiu7tTiwy F any f d vector Idea wecanjustcheckthisfoyps.dspauisiso.to 113in Ingen eral the Euclidean 354in Eep We can alwayscomplete T two t.I.co umnvectorsotthesamed.m.asetofL 2 vectorsinV xp column vectors spanks 34 1 2 0 40 10 to a basis u 7.5 pivot columns impossible we can always remove redundancy we can determine pivot columns by tow reductions from aset of spanningvectossofu vi ii will both be pivotal to forma basis
Transcript of W3 - math.berkeley.edu
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 .
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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.
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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?
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