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?

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

?

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 $\displaystyle u\notin A^{\perp}$?

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 $\displaystyle A \subseteq B$.

re: orthogonal subset proof

aahhhh correct

i understood this point

now i think about the profe

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?