QUESTIONS-ANSWERS-LINEAR ALGEBRA
QUESTIONS/ANSWERS... BY ACF - LINEAR ALGEBRA
Feb.6 th.(12 hours ago)
acf has left a new comment on your post "QUESTIONS - CDOG - MATRICES":
If A1, A2,...AK which is an element of R to the nth dimenion are non-zero and mutually orthogonal show that they are linearly independent.
Posted by acf to Maths,management,counselling at 7:06 AM
FOR EASE OF NOMENCLATURE LET ME CHANGE THE PROBLEM AS FOLLOWS.
A,B,C,........K ARE NON ZERO,MUTUALLY ORTHOGONAL,N DIMENSIONAL VECTORS.
TST A,B,C ......K ARE INDEPENDENT.
LET US PROVE THIS BY REDUCTIO-AD-ABSURDUM.
THAT IS BY CONTRADICTION.
LET US ASSUME,A,B,C.....K ARE DEPENDENT.
THEN WITHOUT ANY LOSS OF GENERALITY, WE CAN TAKE THAT
A=X2*B+X3*C+X4*D+.............Xk*K......................1
DOTTING IT WITH B,C,ETC...AND NOTING THAT THE VECTORS ARE MUTUALLY
ORTHOGONAL THAT IS A.B=A.C=...=B.C=B.D=.....0...WE GET,
A.B=0=X2(B.B)+X3(C.B)+.......Xk(K.B)=X2(B.B).....2
A.C=0=X2(B.C)+X3(C.C)+.......Xk(K.C)=X3(C.C)......3
...........................................ETC
A.K=0=X2(B.K)+X3(C.K)+.......Xk(K.K)=Xk(K.K)......4
---------------------------------------------------------------------------
SINCE B,C.....K ARE NON ZERO VECTORS ,
B.B,C.C,........K.K CANNOT BE ZERO .
HENCE X2=X3=......=Xk=0
THAT IS
A=0...........FROM EQN.1
BUT THIS A CONTRADICTION SINCE A IS NON ZERO VECTOR.
SO OUR ASSUMPTION THAT
A,B,C,......K....ARE LINEARLY DEPENDENT IS WRONG.
HENCE A,B,C.....K ARE LINEARLY INDEPENDENT.
posted by mathsmanagementcommonsense at 8:35 PM
4 Comments:
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2:50 PM
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8:02 AM
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