```Question 33286
Suppose S,T are subspaces of V and s intersect T=0.Show that every vector in S + T can be written uniquely in the form A + B, with A in S, and B in T.

THAT IS S AND T ARE DISJOINT..THEY HAVE NO ELEMENTS IN COMMON.SINCE THEY ARE DISJOINT SUBSPACES OF V ,THEIR DIMENSIONS ARE LESS THAN THAT OF V AND THEIR BASIS' SHOULD BE A PAIR OF DISJOINT SUBSETS OF THE BASIS OF  V WHATEVER MAY BE THE BASIS THAT HAS BEEN ADOPTED FOR V
SAY IN R3,IF ELEMENTS OF S ARE (X,0,0)THEN ELEMENTS OF T CAN BE (0,Y,0)..OR (0,0,Z)...OR...(0,Y,Z)...OR SO...
SAME APPLIES TO ANY DIMENSION OF R LIKE R4,R5...RN.
SO LET US TAKE S WITH ELEMENTS S1,S2...ETC AND T WITH ELEMENTS T1,T2 ETC.LET THE  BASIS OF V BE B ( B1,B2...BM,BN,BO,BP,BQ,BR).THEN B1,B2,....ETC ARE ALL INDEPENDENT.
IN SUCH A CASE SINCE S IS A SUBSPACE OF V ,ANY VECTOR IN S CAN BE EXPRESSED AS
A UNIQUE LINEAR COMBINATION OF SOME DEFINITE ELEMENTS OF BASIS B....SAY ANY VECTOR IN S BE
S1=(X1B1+X2B2+...XMBM),WHICH IS UNIQUE COMBINATION FOR THIS BASIS, ASSUMING DIMENSION OF S TO BE M , WHERE M<N.....HERE WITHOUT LOSS OF ANY GENEARALITY ,WE TOOK B1,B2..BM TO SPAN ELEMENTS IN S AND BN,BO,BP,BQ,BR..TO SPAN THE DISJOINT ELEMENTS IN T.HENCE
T1=(YNBN+YOBO+YPBP+YQBQ+YRBR)....WHICH IS ALSO AN UNIQUE COMBINATION FOR THIS BASIS
IN FACT AS PER THE THEOREM ON DIMENSIONS OF SUB SPACES,IF V IS A FINITE VECTOR SPACE AND S & T ARE ITS TWO SUBSPACES THEN ,WE HAVE.....
DIMENSION(S+T)=DIMENSION(S)+DIMENSION(T)-DIMENSION(S INTERSECTION T)
SINCE IT IS GIVEN THAT S INTERSECTION T IS ZERO,WE GET....
DIMENSION(S+T)=DIMENSION(S)+DIMENSION(T)....THUS IF V IS OF DIMENSION R , AND ITS SUBSPACE S IS OF DIMENSION M THEN DIMENSION OF SUBSPACE T WOULD BE LESS THAN OR EQUAL TO R-M,EQUALITY BEING WHEN S AND T TOGETHER SPAN THE ENTIRE VECTOR SPACE OF V.
HENCE ANY VECTOR IN S+T....THE UNION OF S AND T WILL BE SAY...S1+T1=(X1B1+X2B2+...XMBM)+(YNBN+YOBO+.....)..AS S1 AND T1 ARE DISJOINT.LET THIS BE =A+B..AS MENTIONED IN THE PROBLEM,BUT A BEING EQUAL TO S1 ONLY AND B BEING EQUAL TO T1 ONLY.
WHICH IS AN UNIQUE LINEAR COMBINATION OF ONE VECTOR IN S AND ANOTHER VECTOR IN T.
THUS ANY/EVERY VECTOR IN S+T CAN BE WRITTEN UNIQUELY AS A+B WITH A IN S AND B IN T.

Construct an example to show that this is false if S intersect T does not equal 0.
NOW WE ARE GIVEN THAT S AND T ARE NOT DISJOINT OR THERE ARE SOME COMMON ELEMENTS BETWEEN THEM.SO FOR AN EXAMPLE LET US TAKE R3 AND ELEMENTS OF S AS S=(X,Y,Z) AND ELEMENTS OF T AS  T=(X,0,0)....NOW TAKE ANY ELEMENT
S1 IN S BE SAY (3,2,1)..THIS CAN BE WRITTEN AS
S1=(3,2,1)=3(1,0,0)+2(0,1,0)+1(0,0,1)..WHICH IS UNIQUE..SIMILARLY
T1=(5,0,0)=5(1,0,0)...WHICH IS UNIQUE....NOW WE HAVE
S1+T1=(8,2,1)=8(1,0,0)+2(0,1,0)+1(0,0,1)WHICH IS UNIQUE COMBINATION WITH THE ADOPTED BASIS.SO WE HAVE S1+T1=1*S1+1*T1..A LINEAR COMBINATION OF ONE VECTOR IN S AND ANOTHER IN T SAY S1(3,2,1) AND T1(5,0,0).....BUT WE CAN EASILY SEE THAT THE SAME (8,2,1)=1S1+1T1
CAN ALSO BE WRITTEN AS A LINEAR COMBINATION OF
S1'=(0,2,1) AND
T1'=(8,0,0)...THAT IS....1S1'+1T1'.......ETC..IN FACT WE CAN GIVE MANY POSSIBLE COMBINATIONS HAVING (X,Y,Z) FORMAT IN S AND (X,0,0) IN T .THIS IS HAPPENING BECAUSE S AND T ARE NOT DISJOINT.WE COULD NOT HAVE DONE THIS IF THEY ARE DISJOINT.
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THOUGH THIS IS NOT ASKED FOR, I AM GIVING IT TO SHOW THE DIFFERENCE.IF S AND T ARE DISJOINT...
SAY IF S IS OF TYPE (X,0,0) AND T OF TYPE (0,Y,0)..THEN LET
S1=(5,0,0)=5(1,0,0)
AND T1=(0,3,0)=3(0,1,0)
THEN S1+T1=(5,3,0)AND THIS CAN BE WRITTEN ONLY AS
5(1,0,0)+3(0,1,0)=1S1+1T1..ONLY```