WHY or HOW , in R3, is (a,b,c) where b=a+c, NOT A VECTOR SUB- SPACE?
WHEREAS in R3, (a,b,c) where c=a-b IS A VECTOR SUB SPACE?

Question

WHY or HOW , in R3, is (a,b,c) where b=a+c, NOT A VECTOR SUB- SPACE? 
WHEREAS in R3, (a,b,c) where c=a-b IS A VECTOR SUB SPACE?

thomas5267 · Answer

@ganeshie8 I am currently studying linear algebra but I have no idea what is happening here.

anonymous · Answer

YEAH, man, I am trying to understand also, because the first one, according to me , is closed under addition, i tested with some examples such as v1= ( 2,5,3)  + v2= (5,9,4) to where v1 and v2 are vectors in V a subset of W, of the form (a,b,c) , so v1 + v2 = (7,14,7) which still suit the given operation b=a+c, same with 3(2,5,3) where 3 is scalar k, which gives me (4,10,6) .

thomas5267 · Answer

If it is closed under addition and it is not a linear subspace, then it must be open under scalar multiplication.  I don't see how it is opened though.

anonymous · Answer

i don't really know, honestly

thomas5267 · Answer

Are you sure that the b=a+c set is not a linear subspace?  Who told you that it isn't a linear subspace?

anonymous · Answer

its in the book, ELEMENTARY LINEAR ALGEBRA, Howard Anton ex 4.2 number 1c

thomas5267 · Answer

I suspect something is wrong with the book.

anonymous · Answer

i dont eve understand it either

anonymous · Answer

@MERTICH I suspect it's a typo as well - looks like a subspace to me.

A note on your work so far: showing that the rule holds for a few examples doesn't establish that the vector is a subspace (unless you find a counter-example, then you only need one to show it's not). Work with the general form.

Take two vectors, $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$, then adding them yields $(a_1+a_2,b_1+b_2,c_1+c_2)$, which according to the rule is equivalent to $(a_1+a_2,(a_1+a_2)+(c_1+c_2),c_1+c_2)$. Hence the set of vectors is closed under addition.

Now take one vector, $(a_1,b_1,c_1)$, and multiply by a scalar $k$. You get $(ka_1,kb_1,kc_1)=(ka_1,k(a_1+c_1),kc_1)$. Hence the set is also closed under scalar multiplication.

Another condition you must satisfy is that the set must contain the zero vector, which occurs for $a=c=0$. You could claim that the set isn't a subspace if $a,b,c$ must be non-zero or distinct, but if that were the case then the second set wouldn't be a subspace.

anonymous · Answer

thank you @sithsandgiggles, and @thomas5627. You have helped me to confirm my sentiments as well