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

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

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

-

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

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

-

5

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

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

-

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