# orthogonal subset proof

• Jun 24th 2011, 12:17 PM
transgalactic
orthogonal subset proof
A,B are subsets of$\displaystyle R^{n}$prove that if$\displaystyle A^{\perp}\subset B^{\perp}$then there is $\displaystyle v\epsilon A$ so v is not part of SpB
how i tryed to solve it:
$\displaystyle s\epsilon A^{\perp}$
$\displaystyle t\epsilon B^{\perp}$
if there is $\displaystyle v\epsilon A$ then $\displaystyle v\bullet s=0$ and because $\displaystyle A^{\perp}\subset B^{\perp}$ then $\displaystyle v\bullet t=0$
so because v is othogonal to $\displaystyle B^{\perp}$and to $\displaystyle A^{\perp}$so $\displaystyle SpA\subseteq SpB A^{\perp}=(SpA){}^{\perp}$
• Jun 24th 2011, 11:20 PM
topspin1617
re: orthogonal subset proof
No. You're not correctly grasping the implications of the containment $\displaystyle A^{\perp}\subsetneq B^{\perp}$.

We know that there exists a vector $\displaystyle u\in B^{\perp}\setminus A^{\perp}$. This means that $\displaystyle u$ is orthogonal to every element of $\displaystyle B$, but that there exists an element of $\displaystyle A$ which $\displaystyle u$ is NOT orthogonal to.

Can you prove it from here?
• Jun 25th 2011, 01:27 AM
transgalactic
re: orthogonal subset proof
Quote:

Originally Posted by topspin1617
No. You're not correctly grasping the implications of the containment $\displaystyle A^{\perp}\subsetneq B^{\perp}$.

We know that there exists a vector $\displaystyle u\in B^{\perp}\setminus A^{\perp}$. This means that $\displaystyle u$ is orthogonal to every element of $\displaystyle B$, but that there exists an element of $\displaystyle A$ which $\displaystyle u$ is NOT orthogonal to.

Can you prove it from here?

i cant undrstand this term
$\displaystyle A^{\perp}\subsetneq B^{\perp}$
if $\displaystyle B^{\perp}$={(1,0,0),(0,1,0)}
and $\displaystyle A^{\perp}$={(1,0,0)}
then
$\displaystyle A^{\perp}\subsetneq B^{\perp}$
so (0,0,1) orthogonal to $\displaystyle A^{\perp}$ and to $\displaystyle B^{\perp}$
so i cant see how

"but that there exists an element of $\displaystyle A$ which $\displaystyle u$ is NOT orthogonal to."

is possible
?
• Jun 25th 2011, 02:29 AM
topspin1617
re: orthogonal subset proof
Did you read my entire post? Since $\displaystyle A^{\perp}$ is properly contained in $\displaystyle B^{\perp}$, there must exist an element $\displaystyle u$ which is in $\displaystyle B^{\perp}$ but not in $\displaystyle A^{\perp}$. Correct?

Now, consider this vector $\displaystyle u$. What does it mean to say $\displaystyle u\in B^{\perp}$? What does it mean to say $\displaystyle u\notin A^{\perp}$?
• Jun 25th 2011, 05:53 AM
HallsofIvy
re: orthogonal subset proof
Do you see the little "/" in $\displaystyle \subsetneq$ (I didn't at first)? That removes the possibility that A= B.

$\displaystyle A\subseteq B$ means that A is a subset that might be equal to B.

$\displaystyle A\subsetneq B$ means that A is a proper subset of B- it is NOT equal to B.

$\displaystyle A\subset B$ is ambiguous but generally should be considered the same as $\displaystyle A \subseteq B$.
• Jun 25th 2011, 07:32 AM
transgalactic
re: orthogonal subset proof
aahhhh correct
i understood this point
now i think about the profe
• Jun 25th 2011, 07:47 AM
transgalactic
re: orthogonal subset proof
Quote:

Originally Posted by topspin1617
Did you read my entire post? Since $\displaystyle A^{\perp}$ is properly contained in $\displaystyle B^{\perp}$, there must exist an element $\displaystyle u$ which is in $\displaystyle B^{\perp}$ but not in $\displaystyle A^{\perp}$. Correct?

Now, consider this vector $\displaystyle u$. What does it mean to say $\displaystyle u\in B^{\perp}$? What does it mean to say $\displaystyle u\notin A^{\perp}$?

i understand that there are members in $\displaystyle B^{\perp}$ which are not in $\displaystyle A^{\perp}$
so not every vector in $\displaystyle B^{\perp}$ 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($\displaystyle B^{\perp}$)+dim SpB= n
dim($\displaystyle A^{\perp}$)+dim SpA= n
so
dim($\displaystyle A^{\perp}$)+dim SpA=dim$\displaystyle B^{\perp}$+dim SpB
so if $\displaystyle A^{\perp}\subset B^{\perp}$ then dim$\displaystyle A^{\perp}$<dimdim$\displaystyle B^{\perp}$

so dim SpA>dim SpB
what now?