september 29, 2020we.prored the uniqueness 3. let v be a real vector space, and w 1,w 2 are...
TRANSCRIPT
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Week 3
MATH 2040B
September 29, 2020
1 Concepts
1. T : V ! W is a linear transformation if(a) T (v1 + v2) = T (v1) + T (v2)
(b) T (↵v1) = ↵T (v1)
2. Let T : V ! V be linear, then a subspace W ⇢ V is said to beT-invariant if T (W ) ⇢ W . (We will use it later.)
2 Notations
1. N(T ) := {~x 2 V : T (~x) = ~0W } ⇢ V
2. R(T ) := {T (~x) : ~x 2 V } ⇢ W
3. Nullity(T ) = dimN(T ), Rank(T ) = dimR(T )
3 Formula
1. Nullity(T ) + Rank(T ) = dimV
2. Two facts:(a) T is injective , N(T ) = 0 , Nullity(T ) = 0.(b) T is surjective , R(T ) = W , Rank(T ) = dimW .
4 Problems
1. Let T : V ! W be a linear transfomation, and assume that dimV = dimW .Prove that
T is injective , T is surjective
1
dvw are both vectorspace
OTLWJCWTivsw
nn
rnspnf.IONullitycntankcnedimvzdtlznnherriT is injective Nullity ㄒ 0
Randi dimVdim w
T is surjectived he
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2. Let T : R2 ! R3 be given by T (a, b) = (a+ b, 0, 2a� b)a) Show that it is a linear transformation.b) Is T injective?c) Is T surjective?
3. Let T : R ! R be given by T (x) = x+ 1, is it linear?
4. Let T : R2 ! R2 with T (1, 2) = (3, 4), T (3, 4) = (1, 2), what is T (1, 0)?
2
NG P 了
fi憇
位𦲀㠭a D T 122 123 1 ⻔盥b 0 dim RE rankreef _dim R2 0
dim122
a o T VI t 以 ⼆ 7417⼗化 dim pit dimR2⼀⽉ ㄒ a u a T411 T is not surjectiveWe can have to Cal bi bz
Thi bi tlaz.bz ㄒ at az but红Cataztbltbz O 291 tzaz bl.bzal tbl 0 zai bytlaztbz.no292
wrhthrrrrrnn
renren.TLal bi t T az byㄒ ⼜ alibi ㄒ 这an ⼜⼩D 2 a 2bi o 2291 2
b1
2 at bi 0,2G bi
ftㄒ 1 1 1 1 1 3
2 intttthr2OTlltTl ltl ltl4tflltlTl tTll ㄒ 21 年 21 1
3
M泩世 正业 __ ___2 卂 ⼀
l prif l i o 2 1.2 p 3.4we have This gives a system of linear equations
䣎嚠 霝 9發管 启fi 2 1.2 轧怂了 是ㄒ 1.2 1ㄒ 34
1 an assumption that cnn.ir
fislineartranymation 2 3.4
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5. Let T : V ! V be linear,
a) Show that N(T ), R(T ) are T-invariant.
b) Suppose that T 2 = T , show that every v 2 V can be written asa sum v = a + b with a 2 N(T ), b 2 R(T ) , and the expression isunique.
3
I 141⼀⼆NCTIRCTjaresubsPT
yg.BA
Let a ENG then ㄒ a 8 EN T
Let b E RCT then TLb E RCT
iㄧ
T 的 invariant
b E rI.UU.EULE T Tu Tu
b Existence i let UGV.ggSuppose U a b a ENG b
E RCTat b T ⽐ KEV
V a t ㄒ 凹 we let T maps bothsides化 ㄒ a ⼗⼆ 孔
Ta t Ta0 t ㄒ 7
7 TCU ㄒ 2
ㄒ化 ⼆ 化
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4
a V 化 ㄒ a ㄒ4 T 4ˇ
V 4 化 t TU ⼆ 化 T v 0Lrt hd a E NCTNLT RCT
we have showed that There actually exist
i ⼼⼼ ⼭ ⼼5007suppose V a t b a EN CT
b ⼆ TCDD
then ㄒ4 T a t T b
0 t ㄒ T ⽐了
化 bere
To b Y V a t b
i ⼆ a V T
In the end V lTCUDtiftthlwe.pro red the uniqueness
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3. Let V be a real vector space, and W1,W2 are subspaces of V . The sum ofW1 and W2 is defined as
W1 +W2 := {w1 + w2 : w1 2 W1, w2 2 W2}
(a) Show that W1 +W2 is a subspace of V .
(b)If W1 = span (S1) ,W2 = span (S2) , show that W1 + W2 =span (S1 [ S2)(c) Suppose that W1 \ W2 = {0}. Then if R1 ⇢ W1, R2 ⇢ W2are linearly independent subsets, show that R1 [R2 is also linearlyindependent.
4
no
proof i Verify three conditionsDOVEWHW2oitJEWltW2fr0i.jEwHW2
aiEWHW2 frta EF and
Firstly Or Q tou E withIt With
Secondly if I wntwiz.EU it Nzand
wi 品J W21 t W22 Ell tWz品 品
then i 5 Wut WR t Wat W22huntW21st Wiz⼗ W22 E hit Uh
n nW Wz
In the end Let at FUR WE haveˊ
a i a Wut Wiz awDt
WizhEWtWziuNzThus.wecan get Wit Wz is a subspace ofV
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b First we show WitWz C spanSt Uss
For It 式 ⼆ W.tw 2 E W t Wz we have
I ⼆ cnn.azvz ntammnntonnnn.ES pantstbiutbznztn.tbnunnnnrnnrenn.ESPansy
where u ⼀ Um E SI Ui Un ES 2
i.lu Um ui n un3C.SI US 2
i By def we have E E Span cswsz
Next for U EE Span Si US2 we can have
I acu.tn tapUp where Ui Up了 C SI US2
we rearrange all ui.ES and ujasz and if Ui ES 175
we put it in StThen we can get
i 志 ah ⼗点 ajujEW.tw 2
As a result we have W.tw⼆SpanCSIUS is
c Takeany elements from Rl and 122 Forexample u___u n c R
U Um c.kz Consider
志 ai Un⼗点bjvjLet w⼆点 ai Un Z ⼆点bjvj
if We0 then Z then we get
w ⼆ 点 ai Un ⼆0 z 恭bjuj
Ui了 9Vj了 are independentrespectively
i fail and 分了 are all 0
if w z to then we have wzewnwzwspanlfuiDCmZEspanfvjDCWz.scm n m 101 ⼈ we get w z 0 we gobad
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4. Let A 2 Matn⇥n(R), and k be a positive integer such that Ak 6= 0, Ak+1 = 0
(a) Show that�I, A,A2, . . . , Ak
is linearly independent.
(b) Show that�I, A+ I, (A+ I)2, . . . , (A+ I)k
is linearly indepen-
dent.
6
to the first case
As a result we have all the elements from Mumare linearly independent
O
proof Ca considerUr a It a At aKAK 0
Multiply AK a A K 0 toa 0
Multiply A⼼ on both sidesa 1 A t O t 0 ⼆ 0
a 0
repeat Similar steps until we geta o 灬 0 In the end we only haveaKAK 0 aK 0
i We have a ak ⼆ forall of them
b First we introduce the binomial theoremones at by 志别akbmk where 只 1器_i
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7
n n x n_n x ㄨ2 X 1
Now we consider
È ai At I 之⼆0
È ail È问⼼ i贰贰 a 们⼼
意志 a j i switch i j
志 Èai 问 们 0
Fom ca we knowall the coefficientsof Aj
Should be Eu
When Fk
蕊 a i ⾼ Uk 0
when j K 1
点 a i i a ki㑎 tak 㓠Gk l 0 0
ak ie.co
Repeat all these stepsand then we have
all Sai了 are Eu which completes
thisproof