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

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- June 24th 2011, 12:17 PMtransgalacticorthogonal subset proof
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 - June 24th 2011, 11:20 PMtopspin1617re: orthogonal subset proof
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? - June 25th 2011, 01:27 AMtransgalacticre: orthogonal subset proof
- June 25th 2011, 02:29 AMtopspin1617re: orthogonal subset proof
Did you read my entire post? Since is properly contained in , there must exist an element which is in but not in . Correct?

Now, consider this vector . What does it mean to say ? What does it mean to say ? - June 25th 2011, 05:53 AMHallsofIvyre: orthogonal subset proof
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 . - June 25th 2011, 07:32 AMtransgalacticre: orthogonal subset proof
aahhhh correct

i understood this point

now i think about the profe - June 25th 2011, 07:47 AMtransgalacticre: orthogonal subset proof
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?