Determine whether the subset A={(x,y,z):y=x+1 and z=y+1} is a subspace of R3
Solution;
By definition;
W is a subspace if;
1.A "\\neq" "\\empty"
2. A is closed under vector addition.
Test for both conditions;
Clear A is not empty .
Check for vector addition;
If you take arbitrary values ,for example, u= (1,2,3) to satisfy A, since;
y=2=x+1=1+1
z=3=y+1=2+1
Also ,we take another set of arbitrary values which also satisfy A;
v=(0,1,2) "\\in" A;
If we add u and v;
(u+v)=(1,3,5)
From A,
y=x+1 and z=y+1
By using (u+v)
y=1+1=2
2"\\neq" 3
Also ;
z=y+1=2+1=3
3"\\neq" 5
Hence A is not a subspace in R3
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