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put this solution on YOUR website!how do i prove W^(perpendicular sign) is a subspace of R^n
WHAT IS THIS W^(perpendicular sign)?PLEASE CLARIFY OR COPY THE PROBLEM AS IT IS.
IN GENERAL TO SHOW THAT W IS A SUB SPACE , WE TAKE 2 ELEMENTS ALPHA AND BETA FROM W AND 2 SCALARS IN REAL FIELD SAY X AND Y AND SHOW THAT
X*ALPHA+Y*BETA IS ALSO AN ELEMENT OF W..WHICH IS NECESSARY AND SUFFICIENT CONDITION FOR W TO BE A SUBSPACE OF R^N
OK..YOU MEAN SET OF ORTOGONAL VECTORS...
BY DEFINITION, A VECTOR Y IS SAID TO BE ORTHOGONAL TO A VECTOR SPACE VN(R)^M...(MEANING A VECTOR SPACE OF N VECTORS WITH DIMENSION M IN REAL FIELD)IF IT IS ORTHOGONAL TO EVERY VECTOR OF THE SPACE.
LET W BE THE SET OF ORTHOGONAL VECTORS AS GIVEN.LET Y AND Z BE ANY 2 ELEMENTS OF W .IF THEY ARE ORTHOGONL TO VECTOR SPACE SPANNED BY SAY X1,X2,....XN...THEN...ANY VECTOR OF THIS SPAN CAN BE WRITTEN AS
A1X1+A2X2+....+ANXN...AND (A1X1+A2X2+....+ANXN).Y=A1X1.Y+A2X2.Y+.....ANXN.Y=0,SINCE X1.Y=X2.Y=...XN.Y=0.BY DEFINITION OF ORTHOGONALITY AS Y IS ORTHOGONAL AS GIVEN.
SIMILARLY.....(A1X1+A2X2+....ANXN).Z=0
LET US NOW CHECK THIS FOR ANY LINEAR COMBINATION OF Y AND Z ......BY+CZ......WE HAVE
(A1X1+A2X2+.....ANXN).(BY+CZ)=(A1X1+A2X2+.....ANXN).BY+(A1X1+A2X2+.....ANXN).CZ
=B(A1X1+A2X2+.....ANXN).Y + C(A1X1+A2X2+.....+ANXN).Z=0 AS PROVED ABOVE...
SO WE PROVED THAT IF Y AND Z ARE ANY 2 ELEMENTS OF W THEN BY+CZ IS ALSO AN ELEMENT OF W AS WE PROVED THAT IT IS ALSO ORTHOGONAL TO THE VECTOR SPACE OF X1,X2....XN. THIS IS A NECESSARY AND SUFFUCIENT CONDITION FOR W TO BE A SUB SPACE OF R^N.
