mathematics tutorial 2b & linear algebra ii, spring
TRANSCRIPT
Mathematics Tutorial 2b & Linear Algebra II, Spring semester 2021 Midterm panic session, 30th May 2021, 17:00- Today:
Questions? •Midterm preparation •
üüLet 13 12 1 x and EU HHB and Care basesof P
Question If EIBE HIWhat is FBxttt
Why is B a basis
f mit n n b m 2 b Cx
s.a.EE Ä
üÄ iii av FH
Let 13 12 1 x and EIN F P Pi
and care basesof ic
IIIIQuestion If EIBE HIWhat is 3 1 Ftp BB EHC
1 step Find 3 7,3Ehf EF
H 3 1 3 1 y H
3 1 X b t.be Hab AKG2 1 t X
2 step Multiply ftp.EAB
d f EHC
3 Step F 3 11 24 O Gtx
2 X D 2 2
What is Fl b Ickx 1244x D 24H X 13
F b 3 am 2441 34 11 244X 5
13GIB 2 EIN Elb
If D I X üüüü item
What is F D
13 2 4 Elam
ftp.ft.IE
s8oE DB8
eD
change of basis matrix an
S Aß G D
s EMD
II
4 4 49s KöfII
d Cd
Tutorial 3 Exercise
HAN B HK ICH Etc SIEGEF EI si li E
q.iqµ
Etc Sidp
F D F Ä DB EFCD
Hin a L dcalculate spies litt
944 Öl L D IktIii
Checkµ Hit d
ok ELIE A
Linear Algebra II & Mathematics Tutorial 2bNagoya University, G30 Program
Spring 2021
Instructor: Henrik Bachmann
Homework 3: Determinants
•
Deadline: 30th May (23:55 JST), 2021
Exercise 1. (2+3+3 = 8 Points) We define the matrix
A =
0
@3 �1 0
2 0 0
�2 2 �1
1
A
and the polynomial
P (�) = det(A� �I3) ,
where I3 denotes the 3⇥ 3-identity matrix.
i) Calculate det(A).
ii) Find all solutions to P (�) = 0.
iii) For each solution � in ii) find a non-zero vector v 2 ker(A� �I3) and evaluate Av.Can you observe a relationship between v, � and A?
Exercise 2. (2+4 = 6 Points) (Geometric interpretation of the determinant)
We define the vectors v =
✓4
2
◆, u =
✓2
3
◆2 R2
.
i) Connect the endpoints of the vectors 0, v, u and v+u to get a parallelogram in R2. (Make a sketch)
ii) Show that the area of this parallelogram is given by det
✓4 2
2 3
◆, i.e. the determinant of the matrix
which has v and u as columns.
(Bonus: Show that this works in general, i.e. if you write two vectors in R2into the columns of a matrix
A 2 R2⇥2then | det(A)| gives the area of the parallelogram spanned by them.)
Exercise 3. (6+2 = 8 Points)
i) Show that the determinant is linear in each row, i.e. for any A = (ai,j) 2 Rn⇥nand 1 l n show
that the map
FA,l : Rn �! Rx 7�! det(A(l;x))
is linear. Here A(l;x) denotes the matrix A, where the l-th row is replaced by the vector xT.
(See at the bottom of page 7 in the overview notes)
ii) Assume that A is invertible. What is the kernel of FA,n?
Version: May 13, 2021- 1 -
einKerl spankB
dXERYBKO
X 14 0
91218 KH o_O
x t
XEtX
tfytekkerlFA.n
fxe.IR detlAlniHIO
rir.euEEnEIXrn
Linear Algebra I - Midterm preparationNagoya University, G30 Program
Spring 2021
Instructor: Henrik Bachmann
Possible midterm type exercises from the homework: HW1 Ex. 1, Ex. 2, HW2 Ex. 2.
1) Decide if the following statements are true or false. Justify your answer by giving a short explanation.
i) The set U = {f 2 C1(R,R) | f 00
= 2f} is a subspace of C1(R,R).
ii) The set U =�A 2 R2⇥2 | det(A) 6= 0
is a subspace of R2⇥2
.
iii) If (b1, b2) is a basis of a vector space V , then (b1 + b2, b2 � b1) is also a basis of V .
iv) If (f1, f2, f3) is a basis of P2, then (f 01, f
02, f
03) is also a basis of P2.
v) For any real number a 2 R there exist at least one linear map F : P1 ! P1 with det(F ) = a.
2) Calculate the determinant of the following matrix
A =
0
@1 2 �3
4 0 �1
1 �2 3
1
A .
3) We define the following elements in P2
b1(x) = x2+ 1, b2(x) = 2x� 1, b3(x) = 1 ,
and define the following function
F : P2 �! P2
p 7�! p0 + 2p .
i) Show that B = (b1, b2, b3) is a basis of P2.
ii) Show that F is a linear map.
iii) Calculate [F ]B .
iv) Determine the determinant of F .
4) Let A 2 R2⇥2and define
UA = {v 2 R2 | Av = 2v} .
i) Show that UA is a subspace of R2⇥2for any A 2 R2⇥2
.
ii) Find A,B,C 2 R2⇥2, such that dim(UA) = 0, dim(UB) = 1, and dim(UC) = 2.
5) Let V be a finitely generated vector space. Show that a linear map F : V ! V is invertible if and
only if det(F ) 6= 0.
6) Show by induction that for all n � 1 we have
nX
i=1
1
i(i+ 1)=
n
n+ 1.
True Ffg fingen amHh
9 Flmüthl
det B 6 1,3EinixD F x 3
man 2mxth This is Suri because
for any axtbePwe have FLEX
aTherefore im F P
F V Wisom dimlvtdin.tw
injective Kerlsurjective im Fw
I
False because Uderzo
TrueFF False f f f Iii
True
1 f GE U f Zf g 2g
f g f g 2ft 2g 2 f tg
ft gehtiii Want to show spank aß_ V
We have BE IIII since for any HV7kHz k X.bt kbzX.at kaFnfz I a
tfIHazvaElRF P P
mxxbmamxtbFHqxttdetttt.de
HEIß EHE13 4,13 ÄH tot
bi
F b Nx D
XX
B Bz
EisOk 444
flxtaxtbxtc ab.CH tE YtfatktfbslHFor any a he ER aha achte
buhlt span Pedi b bz is a basis
3 3
ii Try yourself ER
Iii E BÄhm ÄH EY l 0L
f bb 2 04 0426
FCB 2 2441 2 42 2
2bkxt b.CH bsCx
2 42 2 1
EMD g
ich detE detfEH 2kz
perf F p 75Ay Zu
A Killer
i OEUA B 3 C 8
KWEUA tAv 2v.Aw
2wAlvtwI_AvtAw_ 2vt2w_ 2lutwI
9 U gWE UA
Alcatel a
Ar zu Edit v
lacht old30
µ6jo
I4 spankB II 8
dinkel