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Show that the set of all vectors (x,y,z) such that x+y+z=0 is a subspace of R3. Solution Show that the set of all vectors (x,y,z) such that x+y+z=0 is a subspace of R3. THE NECESSARY AND SUFFICIENT CONDITION FOR A SET S TO BE SUB SPACE OF V IS THAT ........ S3=P*S1+Q*S2 SHALL BE AN ELEMENT OF S ....WHERE ..... S1,S2, ARE ANY 2 ELEMENTS OF SET S AND P AND Q ARE ANY 2 SCALARS..... LET S1=[X1,Y1,Z1]..........................S2=[X2,Y2,Z2] .......THEY ARE ELEMENTS OF S ...SO AS PER GIVEN CONDITION , WE HAVE ... X1+Y1+Z1=0........................................1...................... X2+Y2+Z2=0........................................2........................... S3=P*S1+Q*S2=P[X1,Y1,Z1]+Q[X2,Y2,Z2] = [PX1+QX2 , PY1+QY2 , PZ1+QZ2] .. SO NOW LET US CHECK IF PX1+QX2 + PY1+QY2 + PZ1+QZ2 = 0 OR NOT .... PX1+QX2 + PY1+QY2 + PZ1+QZ2 =P[X1+Y1+Z1]+Q[X2+Y2+Z2]=P*0+Q*0=0 SO................. S3=P*S1+Q*S2 IS AN ELEMENT OF S ........... SO S FORMS A SUB SPACE ..........PROVED ............................ .

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