SOLUTION: Let u={(x1,x2,x3)in R3|x1+x2+x3=a} for a in R3 fixed. Show the u is a vector subspace if and only if a=0. This abstract and I'm struggling on the abstract ideas in linear algebr

Question 33284This question is from textbook Linear Algebra
: Let u={(x1,x2,x3)in R3|x1+x2+x3=a} for a in R3 fixed. Show the u is a vector subspace if and only if a=0.
This abstract and I'm struggling on the abstract ideas in linear algebra.
Thank you for any help you can give me!!! This question is from textbook Linear Algebra

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
BEFORE I ANSWER THIS LET ME ASK YOU ,YOU TOOK THIS
ABSTRACT ALGEBRA BY YOUR CHOICE OR IT IS COMPULSORY IN YOUR COURSE OF
STUDY.WHY I AM ASKING YOU THIS IS BECAUSE,YOU SAY YOU ARE STRUGGLING WITH THIS ABSTRACT COURSE AND INDEED THIS IS A DRY AND HIGHLY
THEORETICAL SUBJECT AND UNLESS ONE HAS AN APTITUDE FOR THIS
THEORETICAL STUDY , ONE IS LIKELY TO GET INTO PROBLEM IN ANSWERING
THESE QUESTIONS.SO FAR THESE ARE LITTLE ELEMENTARY PROBLEMS BUT IF THEY
GET INTO ADVANCED TOPICS ,IT WILL BE MORE DIFFICULT.A COURSE IN
APPLIED MATHS ON THE OTHERHAND WOULD BE EASIER IN GENERAL.FOR MANY NOT
INCLINED FOR THEORETICAL ABSTRACTS.
OK LET US GET ON
THE NECESSARY AND SUFFICIENT CONDITION FOR U TO BE A SUBSPACE IS THAT IF P AND Q ARE ANY 2 ELEMENTS OF U,AND C AND D ARE ANY 2 ELEMENTS IN R3 THEN CP+DQ SHOULD BE AN ELEMENT OF U.
SO LET P (X1,X2,X3)AND Q (Y1,Y2,Y3)BE 2 ELEMENTS OF U.HENCE WE HAVE FROM THE GIVEN CONDITION
X1+X2+X3=Y1+Y2+Y3=A......................................I
THEN....CP+DQ =C(X1,X2,X3)+D(Y1,Y2,Y3)=(CX1,CX2,CX3)+(DY1,DY2,DY3)
=[(CX1+DY1),(CX2+DY2),(CX3+DY3)]
NOW FOR THIS TO BE IN U ,IT SHOULD SATISFY THE GIVEN CONDITION .....NAMELY..
CX1+DY1+CX2+DY2+CX3+DY3=A
C(X1+X2+X3)+D(Y1+Y2+Y3)=A...FROM I....WE GET
CA+DA=A
A(C+D)=A...................NOW SINCE C&D ARE ANY 2 ELEMENTS IN R.....C+D CANNOT BE TAKEN AS EQUAL TO 1.HENCE THIS CAN BE SATISFIED IF AND ONLY IF A=0
FOR IF A=0 THEN WE HAVE 0*(C+D)=0..HENCE U IS A SUBSPACE.
IF A IS NOT EQUAL TO ZERO,WE SHOWED ABOVE THAT U IS NOT A SUB SPACE.
HENCE U WILL BE SUBSPACE IF AND ONLY IF A=0
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