Διάλεξη 27η - Προβολές και ελάχιστα τετράγωνα

GRAMMIK'H 'ALGEBRA27h di�lexh:Probolèc

23 NoembrÐou 2012

Probol  se eujeÐa tou Rn

Na brejeÐ h probol  p tou b ep�nw sthn eujeÐa

pou orÐzei to a

mNa brejeÐ to plhsièstero sto b shmeÐo p thc

eujeÐac pou orÐzei to a

To a dièrqetai apo thn arq  twn axìnwn.

Probol  se eujeÐa tou Rn

Probol  se eujeÐa tou Rn

H probol  p enìc dianÔmatocb ∈ Rn se mia eujeÐa a ∈ Rn poupern�ei apo to 0

- eÐnai h p = aTb

aT aa

- me antÐstoiqo pÐnaka probol c

P = aaT

aT aI που είναι συμμετρικός και τάξης 1

H probol  p enìc dianÔmatocb ∈ Rn se mia eujeÐa a ∈ Rn poupern�ei apo to 0

- eÐnai h p = aTb

aT aa

- me antÐstoiqo pÐnaka probol c

P = aaT

aT aI που είναι συμμετρικός και τάξης 1

H probol  p enìc dianÔmatocb ∈ Rn se mia eujeÐa a ∈ Rn poupern�ei apo to 0

- eÐnai h p = aTb

aT aa

- me antÐstoiqo pÐnaka probol c

P = aaT

aT a

I που είναι συμμετρικός και τάξης 1

H probol  p enìc dianÔmatocb ∈ Rn se mia eujeÐa a ∈ Rn poupern�ei apo to 0

- eÐnai h p = aTb

aT aa

- me antÐstoiqo pÐnaka probol c

P = aaT

aT aI που είναι συμμετρικός και τάξης 1

Par�deigma

- Probol  tou

123

sto

111

- PÐnakac probol c sto

111

kai

sto

[cos θsin θ

]

Par�deigma

- Probol  tou

123

sto

111

- PÐnakac probol c sto

111

kai

sto

[cos θsin θ

]

Par�deigma

- Probol  tou

123

sto

111

- PÐnakac probol c sto

111

kai

sto

[cos θsin θ

]

Efarmog 

UpologÐste x tètoio ¸ste Ax = b kai x /∈ R(A).

Efarmog 

UpologÐste x tètoio ¸ste Ax = b

kai x /∈ R(A).

Efarmog 

UpologÐste x tètoio ¸ste Ax = b kai x /∈ R(A).

Efarmog 

UpologÐste x tètoio ¸ste Ax = b kai x /∈ R(A).

El�qista Tetr�gwna

An Ax = b kai x /∈ R(A) tìte

miaprosèggish thc lÔshc x eÐnai hlÔsh y tou sust matoc Ay = pìpou p h probol  tou b ston R(A).

El�qista Tetr�gwna

An Ax = b kai x /∈ R(A) tìte miaprosèggish thc lÔshc x eÐnai hlÔsh y tou sust matoc Ay = pìpou p h probol  tou b ston R(A).

El�qista Tetr�gwna

Ax = b

⇒ ATAx = ATb⇒

x =(ATA

)−1ATb

Pr�gmati AT (Ax − b) = 0

Upìjesh: oi st lec tou A eÐnaigrammik� anex�rthtec.

El�qista Tetr�gwna

Ax = b⇒ ATAx = ATb

⇒

x =(ATA

)−1ATb

Pr�gmati AT (Ax − b) = 0

Upìjesh: oi st lec tou A eÐnaigrammik� anex�rthtec.

El�qista Tetr�gwna

Ax = b⇒ ATAx = ATb⇒

x =(ATA

)−1ATb

Pr�gmati AT (Ax − b) = 0

Upìjesh: oi st lec tou A eÐnaigrammik� anex�rthtec.

El�qista Tetr�gwna

Ax = b⇒ ATAx = ATb⇒

x =(ATA

)−1ATb

Pr�gmati AT (Ax − b) = 0

Upìjesh: oi st lec tou A eÐnaigrammik� anex�rthtec.

El�qista Tetr�gwna

Ax = b⇒ ATAx = ATb⇒

x =(ATA

)−1ATb

Pr�gmati AT (Ax − b) = 0

Upìjesh: oi st lec tou A eÐnaigrammik� anex�rthtec.

Par�deigma

1 4

1 5

0 6

x =

456