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3406 Lecture 3

GT Math 3406 Second Course of Linear Algebra lecture 3
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0% found this document useful (0 votes)
53 views12 pages

3406 Lecture 3

GT Math 3406 Second Course of Linear Algebra lecture 3
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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LV factorization

1/ Eri I
- 2 o u

A = G

E ,
A
[] EstI=

EE ,
A
= E

E - I

=
F -9
y

*x =
1b =
(2)

I
LUX =
UX =
2
-

Le =
b save for 1 and then

UX =
1 for X .

Le =

(5) = =
()
x = 2
, y
= 0
,
z =
10 L(z)
=
(i) :] =

1)

I (i).

X =

[
Does LV factorization always work !

Example :
-

/100 I
2 12
A = -
32/

Swap
-
rows :

P() (2) =

D =

[
it : 31 ,
2
, .., nb > 51, 2
..., my
It is invertible permutation ③

(123)
23/

I =

( =
/- I
C C

PA =

Perform LV factorization for


this metrix

PA = Lu

=] v
=
AX =
b =>
PAX =
Pb

Lux =
PE
-

PA permutes rows

AP 1)
columns .

[h/[56] =
[I

(AT)ij =

Aji

1: : =

12
=A .

(A + B)T =
AT+ BT

(AB)T = BTAT

(B)
Tij =

B) ji
=
z Aj Bri
k

(T)kj /BT)ik

=
(BT)in
(T)kj
=
(BTAT)
ij

(AT) =
(A-Yi

AA" =
I = (A) =
I

T
=
CA") AT =
I

↑A-)
+
is left inves of AT

AA =
I = (A A)T = I o

A
+
A )T =
I

=>
At T is
right inverse

or (AT) =
<A")
T
.

(Ax)T =
XT AT


AX linear combination of columns. ofK

* TAT linear combinations of

the rows of AT
.

Dot producti

XTAy =

(A)T]

#T
A =
symmetric

A myn
> > At nxm

of IR
-cubspaces
V subspace of IR" .

u ,
veV = A + 1EV

GEV ,
J Scalar = JEV .

Us ,
Unc-- > AktV C . -

, 2 Scalers
K

Ester
Amxn
·
CLA) column space is a subspace

of IRM

· NCA) =
SEIR" : Ax =
03 ,

NCA) is a subspace of IR".

*,, Xz N(A)

A(X, 12)
#+
=>
+
&
= =
.

O
-
G
↓ ENIA) a scalar .

↑ (1)
d Es
=

I e
12

A =

I
1 -
1 -
12

/2 J I ·
01 2 ( C 12

, 2 C
-> - O I
O 12 C C O C
C C 00
0 -
1 -
2 C

Ro

() (
2z 0x + z + zw= 0
y + =

z(-2) w() +

1/
special solutions .

besis for
they are
a

me
hull space.

Basis V subspace

V
-
1 . .... Up is a beas if

Ev were ex <, --
E with
.
,
k

= Es
j = 1
,
i . e
. , .... Uk

Spen .
V

1. -- > As are .
knearly independent

The special solutions of Ax =*

form a basis for NCA)


.

)N(Rd) =

not
solution space does change
under now reduction
.

CCA)
- ((Ro)

I
-

I
11 2

I
O
10 12

I
2 120
A =
C ↓ zu R
j =

O 00 O
1 32 O oo G

L -
1 -
12
Ro :
In ⑤

1 2(g)-1:1 /:+ =

Ro() 1:↑ =

A
(=2) v, +
24 1 g
-
=
=

24
+
-

A-1, o ↓
2

&, &2 , As &4

&, 82 , &3 =
&, +
242

&4
= 29 ,

A =
[a ,
a) Id ? If
C R

It will turn out that in the

CR factorization R will be

the reduced echelon form

The number of pirchs =


# of

knearly indep columns that

Spen CCA) .

# of free veviables =
# of

special solutions
. # of
columns
.

#of Hirots + # of special sol .
=

dimCCA) + chimNCA) =
.
n

Fundamental tim of him .


algebre
dimCCA) = renk of A
i 114
(1 , 92 , 13 &4)
,

I
A, diz 93 Alt
an
031
C+
d2z

Az AAs
Ays 924

a+ 4
I

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