ÉIiIÉYi
sirf
and
has
one
Choose
to do
arbitrary
an
to solve
Try
Two
the following
fix
a e
no
4
1 4
cases
This equation has
unique
a
solution x v e V
for all way
Then the
required g exists and
it takes a
v
The equation either has no
solution for at least one
vector u EU or it has
multiple solutions
Then
g doesn't
not invertible
Example
f
exist
123
P2
pan
Is f invertible
Sm
formula
Q
p x
PC s
p
p
a
o
s
a
tb
x
a
b te
a
at
exam
Take
u
Need
a
an
YA
etc
Inc
f is
I
ex 2
b te
fi
c
arbitrary
e
Pas
vector
123
E3
Ey
Linear system with respect
I
I
a
reduced
echelon form
92
fact 93
9 92 21
93
6
C
Hence the inverse
f
a b c
is
map
given by
R
ft 123
az
ag
ta az
notation for the inverse
map
taraz g x
x
Another
way to determine if a LM
invertible or not is to look
at
its matrix
th
let f
and apply the
U be
V
linear map
Let B V
C C U be bases
Then f is invertible
iff
is invertible the last
ate
Prod
Btc
det
happens iff
I
a
1 In
0
In
1
invertible Want
is invertible
Given f is
Cfc
Since f is invertible
sit
Remind
Id
got
EE
II
In
B
B
n
LI JEE
to
F g ill
got
fee
Ig
dint
In
E
I
By def
its
he
is
Cfc
Want to
EB
is
inverse
Given
invertible and
is
show
invertible
that f is invertible
fue 4
equivalently
fax u has a unique solution
jestificatif
comes
i.e f is injective and surjective
later
f is injective if f
We know
today
We proved before
kerf 0
that elements in the kernel corresponds
or
to
elements
But
Etc
Hence
Late
kerf
Iiit
03
03
t is surjective
Inf
Since
is
a
if f
dim Imf
diary
need to prove
that
dim In f n
Apply nullity rank Th
dim kerf dim In f dim
dim kerf 0
Hence dim Imf
A
4
Example
f
Pz
ff
vector subspace
this happens
we
i
U
Inf
So
Cfc
invertible
is
Nal
Hence
Nal
in
Ma IR
V
g
pix
Pez
Q
Is f invertible
P 1
Solution
pox
PC 2
p
a
task 192
90 29,4692
1
493
I
3
893
92 292 39 X2
a
P
pls
P a
1
2924393
292 603 X
I
693
693
292 693
P
f
2
got a xtqx
2
93 3
291 492 893
292 693
92 292 393
fo
13
693
I x X X
C
Late
en en
1
f 1
x
Jc
711
88
fix
81
OTS Xt O x2 0 x
Ease
1
1
11
Io
1
0 x 0
2
0 3
la
ez
E
1 1 2.6
fd
det
Matrix is invertible
f
is
By the thereen
invertible
U be
t
the let t
1240
LM
a
f is invertible i f f
f is injective and surjective
Then
Proof
Given f is
invertible
T sit
I g U
fog
got Id
Want
f is injective
surjective
Injectivity
Take
go f
Need
ve
v
other hand
go f Id
go f v
follows that
It
Surjectivity
F
v
EV
at
flu
ay
Done
O
et
Mstake O into 0
o
v
v
0
as
any
E
u
Iffy
Need
f
o
g
Idu
for
sit
0
d
On the
Hence
and
Then
g C
u
Idy
kerf
kerf
i.e
fog
required
Y
flay
Idy a
u
4
i
Given
f is injective and surjective
construct a l m
V sit
U
Want
g
Take
Want
EU
a
vector
a
look
f x
to associate
T
in
Id
got
fog Ida
call
yell
U
f is surjective it has a
solution
Since f is injective a solution
is unique
let y be the solution
since
so
construct
we
t
U
g
a
4
map
Ty
It remains to
prove
linear
of
Need
Idu
dog
42,4 E
U
g 41 42
let
Vy
Vaz
f Vas
9 us
LHS
g us
Uf
Uz
vastVue
compute
Observation
f
Idg
the solutions
of
f Vaz
RHS
and
map
is a
g
since
rust f Va
11
41442
g
Usta
f is
linear
f Mutua
So vastly
of
f
solution
a
is
7 41 42
x
So LHS
usta
g
VastVaz
To sum up
we proved
that
g respects addition
for
Similarly
Example
EF
MEU
a
P
V
B
for all
gcu
a
glance
I
x
all polynomials
42,43
3 basis
dim P
at
UCP
I
as X
y
t
03
1103
I
azx't
is
Consider
an X
in
proper
D
f
4
pox
Q
A
P
xp ex
Well defined xpex has no
constant coff
Determine if
f is invertible
need to check if f is
injective and surjective
injectivity
Take
Pax
pix
need
e Ker f
Q
X pix
kerf
O
This happens iff
0
Pex
so
Ker 7 0
surjectivity
Take any
apt 92
Need
21
find pox
Take pix
a
U
Tan XU
9 X Taz Xt
f pas
X
in
vector
9
d
9 X t 9242t
XU
anXh l
az Xt
tazXt
ta
tan yay
9 a Xh
So f is surjective
hence f is invertible