&DStWXOR±$SOLFDo}HV/LQHDUHV47
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Å
(VSDoRV ,VRPRUIRV
x
Sejam ( e (¶ dois espaços vectoriais sobre £.
Dizemos que ( e (¶ são LVRPRUIRV se H[LVWLUXPLVRPRUILVPR M
e escrevemos,
x
: ( | (¶
( ! (¶
Por exemplo,
o espaço dos vectores livres no plano é LVRPRUIR a ¸ ,
2
porque podemos estabelecer um LVRPRUILVPR
que, à extremidade de cada vector, faz corresponder
as respectivas coordenadas [\ ± ¸ .
2
x
3URSULHGDGH: Sejam ( e (¶ dois espaços vectoriais sobre £,
se
então
x
M : ( | (¶ é um LVRPRUILVPR
M-1: (¶| (
também é um LVRPRUILVPR.
3URSRVLomR: Sejam ( , (¶ e (¶¶ WUrVHVSDoRVYHFWRULDLV sobre £,
D (
!(
E se (
F se (
! (¶então (¶! (
! (¶e (¶! (¶¶então ( ! (¶¶
ou seja, o LVRPRUILVPR é uma UHODomRGHHTXLYDOrQFLD.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV48
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
'HPRQVWUDomR: D
(!(
A DSOLFDomRLGHQWLGDGH
que a todo o X
LG( : ( | (
± ( faz corresponder LG( X X
é uma aplicação linear, injectiva e sobrejectiva,
logo um LVRPRUILVPR.
E se (
! (¶então (¶! (
Se existe um isomorfismo
M : ( | (¶
então, pela propriedade anterior, a DSOLFDomRLQYHUVD
M-1: (¶| (
F se (
WDPEpPpXPLVRPRUILVPR.
! (¶e (¶! (¶¶então ( ! (¶¶
Se existe um isomorfismo
e um isomorfismo
M : ( | (¶
\ : (¶| (¶¶
então, a DSOLFDomRFRPSRVWD
\ Ì M : ( | (¶
é uma aplicação linear, injectiva e sobrejectiva,
logo um LVRPRUILVPR.
x
3URSRVLomR: Sendo ( e (¶ HVSDoRVYHFWRULDLVGHGLPHQVmRILQLWD sobre £,
então,
( ! (¶x
x
GLP(
GLP(¶ .
Este resultado tem consequências práticas particularmente importantes, pois
mostra que todos os espaços vectoriais FRPDPHVPDGLPHQVmRILQLWD são
LVRPRUIRV.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV49
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
3URSRVLomR: Seja ( umHVSDoRYHFWRULDO sobre £ tal que GLP
então,
x
( ! £Q
Por exemplo os HVSDoRVYHFWRULDLV
pois têm GLPHQVmRILQLWD e,
GLP0l(¸)
( Q,
0l(¸) e ¸6 VmRLVRPRUIRV,
GLP¸6
Podemos confirmar este resultado estabelecendo uma aplicação,
I : 0l(¸) | ¸6
e provar que se trata de um LVRPRUILVPR.
Por exemplo a aplicação,
Para provar que é LQMHFWLYD, calculemos o Q~FOHR da aplicação,
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV50
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Assim, como o Q~FOHR de I é apenas o YHFWRUQXOR de 0l(¸),
podemos concluir que é um PRQRPRUILVPR e, tendo os dois espaços
D PHVPDGLPHQVmR, podemos também concluir que é um HSLPRUILVPR.
Falta apenas SURYDU que I é mesmo XPDDSOLFDomROLQHDU ...
x
Por exemplo os HVSDoRVYHFWRULDLV
pois têm GLPHQVmRILQLWD e,
GLP3Q[[]
3Q[[] e ¸n+1
VmRLVRPRUIRV,
Q GLP¸n+1
Efectivamente a aplicação,
M : 3Q[[] | ¸n+1
DQ [Q D [ D DQ D D
é um LVRPRUILVPR de
x
3Q[[] em ¸n+1.
Mostre que os HVSDoRVYHFWRULDLV 3[[] e 0l(¸) VmRLVRPRUIRV,
estabelecendo uma aplicação entre eles e provando que é um LVRPRUILVPR.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV51
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Å
0DWUL] GH XPD $SOLFDomR /LQHDU
x
Nesta secção, todos os espaços vectoriais considerados têm GLPHQVmRILQLWD.
x
Sejam ( e (¶ espaços vectoriais sobre £ tais que GLP(
)
sejam
e seja
)
H H HQ
Q e GLP(¶ S,
uma EDVHRUGHQDGD de (
H¶ H¶ H¶S
uma EDVHRUGHQDGD de (¶
M : ( | (¶ uma DSOLFDomROLQHDU.
A PDWUL]GDDSOLFDomROLQHDU M HPUHODomRjVEDVHV ) e
do tipo S
) é uma matriz
× Q, representada por 0M ) ) e definida por,
onde FDGDFROXQD L Q é formada pelas FRRUGHQDGDV de MHL
na base ),
ou seja,
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV52
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Por exemplo para a DSOLFDomROLQHDU
M[\ [[±\\,
M : ¸2 | ¸3 definida por,
para todo o[\
∈ ¸2
consideremos as EDVHVFDQyQLFDV,
)¸2
)¸3
e calculemos a respectiva PDWUL]
0M )¸2 )¸3.
Determinando as FRRUGHQDGDVGDVLPDJHQV dos vectores da base )¸2 na
base )¸3,
M )¸3
M ±
±
±)¸3
e portanto,
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV53
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Para a mesma DSOLFDomROLQHDU
M : ¸2 | ¸3 definida por,
M[\ [[±\\,
para todo o[\
consideremos agora as EDVHV,
)
)’
±
e calculemos a respectiva PDWUL]
0M ) )’.
∈ ¸2
de ¸
de ¸
2
3
Determinando as FRRUGHQDGDVGDVLPDJHQV dos vectores de ) em )’,
M ±
±)’
M± ±±
±
±)’
e portanto,
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV54
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Por exemplo para a DSOLFDomROLQHDU
IDEF DE±F,
I : ¸3 | ¸2 definida por,
para todo oDEF
em relação às EDVHV,
)
)’
±
calculemos a respectiva 0I
UHODomRjVEDVHV ) e
)¶
de ¸
de ¸
∈ ¸3
3
2
) )’, PDWUL] GDDSOLFDomROLQHDU I HP
Determinando as FRRUGHQDGDVGDVLPDJHQV por I, dos vectores de ) em )’,
I
±
±)’
I
±
±
±
±)’
I± ±
)’
e portanto,
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV55
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Por exemplo para a DSOLFDomROLQHDU
\ : 3[[] | ¸3
\D[ E[F EE±DD
para todo o
em relação às EDVHV,
)3[[]
)
[[ D[ E[F± 3[[].
base canónica de 3[[]
calculemos a respectiva 0\
UHODomRjVEDVHV )3[[] e
)
definida por,
de ¸3
)3[[] ), PDWUL] GD DSOLFDomROLQHDU \ HP
Determinando as FRRUGHQDGDVGDVLPDJHQV por \, dos polinómios de )3[[]
na base ),
\ )
\[ ±
±)
\[ ±
±±
±±)
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV56
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
e portanto,
x
Dada a DSOLFDomROLQHDU
T : 3[[] | 3[[] definida por,
TD[ E[F D±E[F
para todo o
e as EDVHV,
)
)·
x
x
[[ ± [±[
D[ E[F± 3[[].
de 3[[]
de 3[[]
) )·, PDWUL] GD DSOLFDomROLQHDU T HP
UHODomRjVEDVHV ) e )·
calcule a respectiva 0T
Sendo dadas uma DSOLFDomROLQHDU M : ( | (¶ e GXDVEDVHV (uma de
cada espaço vectorial) sabemos como construir XPD~QLFDPDWUL], que
FDUDFWHUL]DHVVDDSOLFDomR.
Deste modo, D DSOLFDomROLQHDUILFDFRPSOHWDPHQWHGHILQLGD se for conhecida
apenas a PDWUL] e respectivas EDVHV.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV57
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
3URSRVLomR: Sejam ( e (¶ HVSDoRVYHFWRULDLV sobre £ e sejam ) e )
EDVHVRUGHQDGDV de ( e de (¶ respectivamente.
e seja M
: ( | (¶uma DSOLFDomROLQHDU tal que,
$
0M ) )
± (,
relativamente à base ) formarem a PDWUL]FROXQD ;,
Se as FRRUGHQDGDVGHXPTXDOTXHUYHFWRU X
então a PDWUL]FROXQDSURGXWR $
FRRUGHQDGDVGH MX±
x
;
(¶relativamente à base ) .
é formada pelas
Retomemos o exemplo da página 54.
Mas desta vez a DSOLFDomROLQHDU
M : ¸3 | ¸2 é GHILQLGDSHODPDWUL],
HP UHODomRjVEDVHV,
)
)’
±
Pretendemos determinar MX, a LPDJHPGRYHFWRU X
de ¸
de ¸
3
2
±±.
Comecemos por calcular as FRRUGHQDGDVGRYHFWRUQDEDVH ).
±± DEF±
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV58
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ou seja,
D ( ... )
DE±F ±
DE ±
D E ±
F ±
e assim,
±± ±±±
±±)
;7
Pela proposição anterior, para determinar as coordenadas de MXna base )·,
basta FDOFXODURSURGXWR $;,
Portanto, M±±
±)·
Por fim, podemos calcular as coordenadas de MX na EDVHFDQyQLFD de ¸ ,
2
M±± ±
±)¸2
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV59
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Retomemos o exemplo da página 53,
com a DSOLFDomROLQHDU
M : ¸2 | ¸3 é GHILQLGDSHODPDWUL],
HPUHODomRjVEDVHV,
)
)’
±
de ¸
de ¸
2
3
Determinemos M e também a H[SUHVVmRJHUDO de M[\,
SDUDWRGR o[\
∈ ¸2
Calculando as FRRUGHQDGDVGRYHFWRU QD EDVH ).
DE±
ou seja,
e assim,
D ±E DE ±)
( ... )
D E ±
;7
o que nos permite determinar as coordenadas de Mna base )·,
FDOFXODQGRRSURGXWR $;,
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV60
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
e portanto,
M )·
)¸3
Para encontrar a H[SUHVVmRJHUDO de M[\, SDUDWRGR o[\
∈ ¸2
o método é análogo. Começamos por,
calcular as FRRUGHQDGDVGRYHFWRUDUELWUiULR[\ QD EDVH ).
[\ DE±
ou seja,
e assim,
D ±E [
DE \
( ... )
D \[
E \±[
[\ \[\±[)
;7
Para determinar as coordenadas de M[\na base )·,
EDVWDFDOFXODURSURGXWR $;,
Portanto,
M[\ \[±\[\)·
que na EDVHFDQyQLFD de
¸3,
\[±\[\
[[±\\)¸3
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV61
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Está assim encontrada a H[SUHVVmRJHUDO,
M[\ [[±\\
SDUDWRGR o[\
∈ ¸2
&RPRDOWHUQDWLYD, partido das FRRUGHQDGDVGH[\ QD EDVH ),
[\ \[\±[)
e conhecendo a EDVH )
±, podemos escrever,
[\ \[\±[±
e, como M é uma DSOLFDomROLQHDU,
M[\ \[M\±[M±
Para calcular
Me M±,
recorremos à própria definição da PDWUL]
0M ) )’.
M ±
M± ±
±±
Portanto,
M[\ \[\±[±±
[[±\\
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV62
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Å
,VRPRUILVPR HQWUH 4( (¶ H 0SlQ¤
x
Sejam
(
e (¶ espaços vectoriais sobre £ tais que GLP
4((¶
0SlQ(£)
x
o conjunto das DSOLFDo}HVOLQHDUHV de ( em (¶
o conjunto das PDWUL]HV do tipo S×Q com elementos de £
Sabemos que,
4((¶
0SlQ(£)
x
( Q e GLP (¶ S.
munido das operações usuais de adição de aplicações e
multiplicação por um escalar, é um HVSDoRYHFWRULDO sobre £
munido das operações usuais de adição de matrizes e
multiplicação por um escalar, é um HVSDoRYHFWRULDO sobre £
Já vimos que, a partir da expressão geral de uma aplicação linear podemos
construir a respectiva matriz, bem como obter a aplicação linear a partir da
matriz.
4((¶ ! 0SlQ(£)
x
O resultado seguinte garante-nos que,
x
3URSRVLomR: Sejam ( e (¶ HVSDoRVYHFWRULDLV sobre £
tais que GLP
( Q
e GLP
(¶ S
sejam ) e ) EDVHVRUGHQDGDV de ( e de (¶ respectivamente
e seja
T : 4((¶ | 0SlQ(£)
Prova-se que
M
TM
T p XPLVRPRUILVPR.
0M ) )
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV63
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Å
([HPSORV GH DSOLFDo}HV OLQHDUHV HP ¹
x
x
Comecemos por considerar o TXDGUDGRXQLWiULR,
delimitado por,
± ¸2
Consideremos também uma DSOLFDomROLQHDU
definida por uma matriz
x
$
M : ¸2 | ¸2
0M )¸2 )¸2,
Calculando as LPDJHQV por M dos quatro pontos que delimitam o quadrado
unitário,
obtemos um quadrilátero, delimitado por,
M M DF
M EG
M DEFG
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV64
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Tratando-se de uma DSOLFDomROLQHDU obviamente que M
x
As duas imagens,
.
M DF
M EG
são as coordenadas das LPDJHQVGRVHOHPHQWRVGDEDVHFDQyQLFD de ¸ ,
2
portanto as GXDVFROXQDV da matriz $.
x
O cálculo da imagem
equivale à VRPD,
M DEFG
DFEG DEFG
x
A LPDJHPSRU M do TXDGUDGRXQLWiULR é portanto um SDUDOHORJUDPR,
x
Calculando o GHWHUPLQDQWH da matriz $ que caracteriza a DSOLFDomROLQHDU,
não é difícil verificar que
_$_ DG±EFé a iUHDGRSDUDOHORJUDPR.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV65
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
Analisemos alguns casos particulares de transformações lineares em ¸ , com
diversas aplicações práticas.
x
A URWDomRSRUXPkQJXOR T, no sentido directo é definida pela matriz,
2
Por exemplo, para T
S ,
M ¹ ¹ M ±¹ ¹ M ¹ Naturalmente que a iUHD do quadrado não se altera pois,
x
Para alterar a área do quadrado, utilizamos a matriz de DPSOLDomRUHGXomR
por um IDFWRU N,
sendo a iUHD resultante dada por _$_
N.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV
&DStWXOR±$SOLFDo}HV/LQHDUHV66
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
x
A matriz de FRPSUHVVmR por um factor N é definida por,
Por exemplo para N
,
M M M o quadrado unitário é comprimido numa das direcções, mantendo a área.
x
A matriz de FLVDOKDPHQWRpor um factor N é definida por,
Verifique que,
M M N
M N
Por exemplo para N
x
,
e muitas mais ...
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
ÈOJHEUD/LQHDU
5RViOLD5RGULJXHV