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Question 34631: I have 2 questions
a/ A is nxn matrix and v1,v2,...,vn be linearly independent in Rn. What must be true about A for Av1, Av2,..., Av3 to be linearly independent?
b/ Must a basis of Pn contain a polynomial of degree K for each k=0,1,...,n ? Justify your answer
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I AM GIVING BELOW A LITTLE MORE ELEGANT SOLUTION TO YOUR FIRST PROBLEM
V1,V2,...........VN...ARE INDEPENDENT
Z1=AV1,Z2=AV2,...........ZN=AVN...WHERE A IS NXN MATRIX
CASE 1. Z,Z2,Z3....ETC...ARE DEPENDENT
HENCE WE CAN FIND K1,K2........KN..NOT ALL ZERO SUCH THAT
K1Z1+K2Z2+...............KNZN=0
=K1AV1+K2AV2+........KNAVN=0
A(K1 V1+K2V2+..........KNVN)=0....................EQN.I
SINCE V1,V2...VN ARE INDEPENDENT AND WE ASSUMED THAT NOT ALL K1,K2....KN...ARE ZERO....HENCE A=0..IF THIS EQN. IS TO HOLD TRUE.
THAT IS A IS SINGULAR MATRIX IF AV1,AV2......AN....ARE DEPENDENT VECTORS.
CASE 2.....SUPPOSE A IS NON SINGULAR.THEN
IN EQN.1,WE HAVE A AS NOT EQUAL TO ZERO AND THE EXPRESSION IN BRACKETS
K1V1+K2V2+....KNVN WILL EQUAL ZERO ONLY IF K1=K2=....KN=0,SINCE IT IS GIVEN THAT V1,V2,....VN ARE INDEPENDENT.THIS MEANS THAT
OUR STRTING POINT K1Z1+K2Z2+.....KNZN=0 IMPLES K1=K2=....KN=0
THAT IS Z1,Z2.....ZN...OR....AV1,AV2.....AVN... ARE INDEPENDENT.
HOPE THIS IS SIMPLER AND EASIER TO PUT UP
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OK GOT IT
OK..LET US USE THE FOLLOWING NOMENCLATURE..
V1,V2,.......VN...FOR N VECTORS..IN R^N.....THEY ARE ACTUALLY.....
V1=(x1,x2,......xN)'....
V2=(y1,y2,......yN)'....ETC..
WE USE THE SYMBOL ' TO INDICATE A TRANSPOSE AS REALLY SPEAKING V1,V2,ETC,,,ARE N ROWS AND 1 COLUMN VECTORS,WHERE AS FOR CONVENIENCE WE ARE WRITING THEM AS 1 ROW AND N COLUMN VECTORS.
A IS A NxN MATRIX SAY..OMITTING THE BRACKETS
a11,a12........a1N
a21,a22........a2N
.......................
aN1,aN2,......aNN.
OR WE SAY A=[aIJ]...WHERE aIJ ARE INDIVIDUAL ELEMENTS OF THE MATRIX.
LET US CALL
AV1=Z1
AV2=Z2
......
AVN=ZN
FROM THE GIVEN RELATION
Z1=AV1=[aIJ]*(x1,x2,......xN)'...
...................................................ETC...
OR IN SIGMA NOTATION
ZI=SIGMA (aIJ*V1) FOR J VARYING FROM 1 TO N.
Z1=a11*V1+a12*V2+...........+a1N*VN
Z2=a21*V1+a22*V2+...........+a2N*VN
......................................................
ZN=aN1*V1+aN2*V2+..........+aNN*VN
WE HAVE TO SHOW THAT Z1,Z2.....ZN ARE INDEPENDENT IF AND ONLY IF A=?,GIVEN THAT V1,V2,.......VN ARE INDEPENDENT.
AS YOU HAVE CORRECTLY GOT THE ANSWER A SHOULD NOT BE SINGULAR FOR THEM TO BE INDEPENDENT.
LET US PROVE IT
Z1,Z2...ZN..WILL BE INDEPENDENT OR NOT DEPENDING ON IF....
K1Z1+K2Z2+.....+KNZN=0....IMPLIES
K1=K2=......=KN=0...OR NOT
AS IS REQUIRED IN IF AND ONLY IF CASES LET US TAKE BOTH THE CASES INCLUDING CONVERSE.
LET US START WITH
K1Z1+K2Z2+.....+KNZN=SAY T IN SHORT = 0..
K1*(a11*V1+a12*V2+...........+a1N*VN)+
K2*(a21*V1+a22*V2+...........+a2N*VN)+
.........................................................
KN*(aN1*V1+aN2*V2+...........+aNN*VN) = 0
SINCE V1,V2......VN ARE INDEPENDENT ,THE EXPRESSIONS FOR Z1,Z2,...ZN INSIDE THE BRACKETS WILL BE ZERO IF AND ONLY IF
ALL OF aIJ ARE EQUAL TO ZERO...THAT IS IF AND ONLY IF MATRIX A IS SINGULAR.
CASE 1 .....MATRIX A IS SINGULAR....
THEN THE EXPRESSIONS INSIDE THE BRACKETS
ARE ALL ZERO...HENCE WE PROVED THAT .
T=O FOR ANY VALUES OF KI...THAT IS T=0 ,BUT IT DOES NOT IMPLY THAT ALL KI ARE ZERO . HENCE
Z1,Z2....ZN ARE DEPENDENT.
CASE 2....MATRIX A IS NON SINGULAR..
THEN THE EXPRESSIONS INSIDE THE BRACKETS
ARE NOT EQUAL TO ZERO..
BUT T=0 .THIS CAN HAPPEN ONLY IF ALL KI ARE ZERO.HENCE WE PROVED THAT IF A IS NOT SINGULAR AND T=0 IMPLIES ALL KI ARE ZERO.
HENCE Z1,Z2....ZN ARE INDEPENDENT..
HOPE YOU UNDERSTOOD THE PROOF .IT IS A LITTLE LEGWORK TO PUT THE FACTS DOWN,ONCE WE ARE CLEAR ABOUT THE PROBLEM AND ADOPT A GOOD NOMENCLATURE.
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I SAW THIS QUESTION EARLIER BUT WAS NOT CLEAR WHAT YOU MEANT...LAST TIME IT WAS MXN MATRIX OF RANK R...ETC...
NOW COMING TO THE PRESENT QUESTION...
PLEASE CONFIRM ...
1.V1,V2,V3....ETC...ARE COLUMNS NUMBER 1,2,3 ETC,,,IN MATRIX A?...THAT IS
A IS GIVEN BY
X1,Y1,Z1...ETC
X2,Y2,Z2..ETC..
X3,Y3,Z3...ETC...
............
XN,YN,ZN...
AND YOU MEAN V1 IS (X1,X2,X3...XN)';{WRITTEN AS TRANSPOSE OF ROW VECTORS FOR CONVENIENCE.
V2 IS (Y1,Y2,Y3.....YN)'
V3 IS (Z1,Z2,Z3....ZN)' AND
V1,V2,V3 ARE ALL INDEPENDENT VECTORS....
2.WHAT DO YOU MEAN BY AV1.....IS IT MATRIX A MULTIPLIED BY V1 VECTOR.N x N
MATRIX MULTIPLIED BY N x 1 VECTOR...
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PART B..
ONCE AGAIN LET US BE CLEAR OF OUR NOMENCLATURE
PI...IS I TH. POLYNOMIAL IN A SET OF POLYNOMIALS
OF THE FORM
PI = aI0+aI1X+aI2X^2+aI3x^3+...............+aINX^N...THAT IS
P1=a10+a11X+a12X^2+a13X^3+...........+a1NX^N
..........ETC......
WHERE X IS CALLED AN INDETERMINATE , aIJ ARE CALLED COEFFICIENTS.tHE + IS NOT TO BE TAKEN AS NORMAL ADDITION AND THE POWERS SHOWN FOR X ARE ALSO NOT NORMAL EXPONENTS ...+ IS USED TO SEPERATE DIFFERENT TERMS IN AN ORDERED SET OF ELEMENTS AND POWERS ARE USED TO INDICATE DIFFERENT PLACES OF ORDER IN WHICH THE TERMS ARE TO BE PUT..HENCE THE POLYNOMIAL CAN VERY WELL BE WRITTEN AS ORDERED SET LIKE
(A,B,C,D...ETC...) WHERE A,B,C,D ARE ALL DIFFERENT ELEMENTS IN THE SET.BUT SINCE IT WAS CUSTOMARY FROM SCHOOLS TO SHOW POLYNOMIALS AS POWERS OF X IT IS BEING CONTINUED HERE.
IF WE UNDERSTAND THIS THEN THESE CAN BE TREATED JUST AS VECTORS EARLIER.
HENCE THEIR LINEAR DEPENDENCE OR INDEPENDENCE FOLLOWS THE SAME RULES.
OBVIOUSLY IN SUCH A CASE IF AN ELEMENT REFERRING TO A PARTICULAR ORDER IS MISSING AS IS THE CASE THAT WOULD HAPPEN IF
BASIS OF PN DOES NOT HAVE ONE OF
degree K for each k=0,1,...,n,THEN IT CANNOT FORM THE BASIS FOR THE GIVEN DIMENSION.AS PER THE THEOREM ON BASIS ,THE NUMBER OF ELEMENTS IN ANY BASIS FOR A SET OF VECTORS (OR POLYNOMIALS HERE BOTH BEING SAME AS MENTIONED ABOVE) IS CONSTANT AND CANNOT BE LESS THAN THE DIMENSION OF THAT SPACE.HAVING ONE OR MORE OF K'S LESS MEANS HAVING A BASIS WITH LESSER ELEMENTS WHICH IS NOT POSSIBLE.
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