# orthogonal subset proof

• Jun 24th 2011, 01:17 PM
transgalactic
orthogonal subset proof
A,B are subsets of $R^{n}$prove that if $A^{\perp}\subset B^{\perp}$then there is $v\epsilon A$ so v is not part of SpB
how i tryed to solve it:
$s\epsilon A^{\perp}$
$t\epsilon B^{\perp}$
if there is $v\epsilon A$ then $v\bullet s=0$ and because $A^{\perp}\subset B^{\perp}$ then $v\bullet t=0$
so because v is othogonal to $B^{\perp}$and to $A^{\perp}$so $
SpA\subseteq SpB
A^{\perp}=(SpA){}^{\perp}
$
• Jun 25th 2011, 12:20 AM
topspin1617
re: orthogonal subset proof
No. You're not correctly grasping the implications of the containment $A^{\perp}\subsetneq B^{\perp}$.

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

Can you prove it from here?
• Jun 25th 2011, 02:27 AM
transgalactic
re: orthogonal subset proof
Quote:

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

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

Can you prove it from here?

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

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

is possible
?
• Jun 25th 2011, 03:29 AM
topspin1617
re: orthogonal subset proof
Did you read my entire post? Since $A^{\perp}$ is properly contained in $B^{\perp}$, there must exist an element $u$ which is in $B^{\perp}$ but not in $A^{\perp}$. Correct?

Now, consider this vector $u$. What does it mean to say $u\in B^{\perp}$? What does it mean to say $u\notin A^{\perp}$?
• Jun 25th 2011, 06:53 AM
HallsofIvy
re: orthogonal subset proof
Do you see the little "/" in $\subsetneq$ (I didn't at first)? That removes the possibility that A= B.

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

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

$A\subset B$ is ambiguous but generally should be considered the same as $A \subseteq B$.
• Jun 25th 2011, 08:32 AM
transgalactic
re: orthogonal subset proof
aahhhh correct
i understood this point
now i think about the profe
• Jun 25th 2011, 08:47 AM
transgalactic
re: orthogonal subset proof
Quote:

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

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

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

so dim SpA>dim SpB
what now?