A,B are subsets of prove that if then there is so v is not part of SpB
how i tryed to solve it:
if there is then and because then
so because v is othogonal to and to so
No. You're not correctly grasping the implications of the containment .
We know that there exists a vector . This means that is orthogonal to every element of , but that there exists an element of which is NOT orthogonal to.
Can you prove it from here?
Do you see the little "/" in (I didn't at first)? That removes the possibility that A= B.
means that A is a subset that might be equal to B.
means that A is a proper subset of B- it is NOT equal to B.
is ambiguous but generally should be considered the same as .
i understand that there are members in which are not in
so not every vector in orthogonal to A
but i dont know how to prove that there is a vector who is in A but not in Sp B
i tried some thing
dim( )+dim SpB= n
dim( )+dim SpA= n
so
dim( )+dim SpA=dim +dim SpB
so if then dim <dimdim
so dim SpA>dim SpB
what now?