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