DIAGONALIZATION/JORDAN FORM Let Ibdarapura/351/lect5.pdf · 2020-04-06 · one-caseapply first hewn...
Transcript of DIAGONALIZATION/JORDAN FORM Let Ibdarapura/351/lect5.pdf · 2020-04-06 · one-caseapply first hewn...
DIAGONALIZATION/JORDAN FORM
Let A = ( Ibd ),
a. b. c. DEIR
del- ( A - XI ) -_ 72- card )x
c- (ad - Sc )-
dotes )=
Use the quadratic formula
X = ¥±Fad2
The number u-her square root iscalled discriminant( a - d )
'
- 4 (ad - Sc )= a'that d ' - 4¥ Gbc
- Z -d
= ( a - d)'
+ Gbc
-
Theorem ②-
(a) if Ca -d)'+45C So ←
fun we two distinct red eigenvalues.
Then A can he dciagonalizealover KR
(b) If ( a - d)"
t 4 be 20then we two distinct complex conjugateeigenvalue . . Then A can be
diagonalized are G .
(c) If ( a - d )'
-14 be = O c-
Hen c) = is i' repeatede.
'
gen value .
I - th 's case,
e. the -
A = ( T o
u x ) or
A is not diagonal . ' really.
REF of co ③
Suppose A has repeated eczema .
I tent A is diagonalize ble .
So A - p ( I g) p - 'for Sue P
.
Then
A = pie ) p- '
= X p z p- '
= > I = ( I ;)Su tf already diagel .
¢We can analyze case ( C )
('
n more detail . Supposedisc = ( a - d)
'
a- 45=0al A is It diagonal , >all
,
Then A does not have ④2 linearly independent eigenvectors
.
Any eigenvector c 's a multiple
of a single eigenvector V,yo
c- ⇒Set N= (A - > I
.
)
= s
)c dWe ha- Null ( N ) e sp - - v
,
lemma NK O-
z
PI N ? IfH. o
O
= o
n)
If we choose#we t a
⑤
multiple af n, ,
then Nr #O
Howe -e - N ( Nr ) = NK-
= o
Thi 's men , Mr c 's an eigen -aah
so Nr = err,
fo - sore
e -40
Replace ✓ by Lev = Vz
aenNIlet P= ( r
, ma ) .
Then
N P = ( Mr.,Nm ) = (o
,
-,)
= cry. ) ( I ! )in
( A - 72 )P= P ( Ed )h⇒=p(o+xp
A f fer e - pending a - I rearranging , ⑧we get
.
There ( Jordan Canon .cl- Fr-- ?Given a hand .
-
a g - nah - zafc,
2×2 red m- Trix A,
then i 's
aninvert . the netrin P
sunt
a. =p ( Y 'a) p - I#
WA -
- C? I'
.
)Diagonalize or put into
Jordan Canal fun .
- ✓atte - simplify
It (A2) = ( x - 1) IoSu D= I is a repented
eigenvalue
So weneed to use T - C. F . ②
first find a nonzero eigenuih
N= A - I = ( i ? )Nunc : )
,
C ' High 19)
⇒ X - y = O
✓ = t ( t ) Choon C- =/
V,I f ! ) is on e.
'
ge -neck
.
W - t re such tht NK -~,
Van. ( '
o ).
( s --In.Cher , ' Ce s )
P -- (nm ) - ( t j )Thu - A =p ) p - '
Check ⑧'
" "÷.
= ( ii. 11%1=1 ? ;D-
-
Application h s > shes fdiff . eg .
¥,
= A ✓,
If A diagrl.i.lyten can solve
If A nut diaguliznll . . Thenuse J .
C. 7 .
A the - a
change f var -allwe p
- '
✓
Get ⑨II. =L : :b
.
w= ( Wiwa ) Then.
dw .If
= A W,c- W,
d.w(I=>€> t
Yoke we = CeeSuh into previous .
and solve.
h set
w,= C
,te> ←
c- ( q - c,)e> t
Then f . .nl ✓ .
-.
There ( Spectre Theurer,
④- re- sea I)tf tf is a red 2×2 symmetric
matrix ( AT -_ A ),
then.
A can be diagonal . -
z -I a-- IR.
rt a. = (ad) evec d - f A
symmetric.
-
- ti 'd )." ,Then the disc - in.int ← squares
is Itai 30And this i 's zero only whena --
d U 5=0.
Then
A- ( g Oa ) d. eh . ;
di - agent .
one- case, apply first hewn frm④
today to .conclude Ai 's
diagonalize -66 on IR.
4ftp.pl.cnt.cn to calculus
. if 2 reefs.
¥)Lef fcx.gl c 's a
"
n .
-
cu"
fl,
if 2 variably ( hi 've mens conf,
-
nun,
& der . In .exist up to 2nd o-les
al are also continuous )
To find naolm.in Sulu- L
3£ -- o, 3dg - o
( et a sul - tu p = ( xo ,yu )called a critical point .
.
Need 2nd der . -hue test . ④
use Hp -
- (3¥.l . II. b)III. I.
'¥1.
By calculus if 's Symmt- is
The 2nd de - - -true test .
Says , he .he- a
c) Local Max at p .
-f the e. jewels
f Hp are both negative
b) Local Min ab p if eigea uh ,
if Hp are both Ros .hu.
c) ( n remain . ycome test
\
'
n Con c ( u g , he .
Ex fcx, y ) = cost cosy
④
what happens at p - CIO) ?
If P c 's g
If I- scarf cosy Cri 'Ll
¥82 - cosxs.az .
} Point22g
= - cosx cosyetc
.
It.- FEI. !
eigenvalues both neg .
So get locntmx .