Question

If M and N are subspaces of a finite-dimensional inner product space, then
\((M+N)^\perp = M^\perp \cap N^\perp\)
and \(M\cap N)^\perp = M^\perp +N^\perp\)

Answer 1

what is \(\perp\) for?

Answer 2

perpendicular

Answer 3

What does \(\cap\) do?

Answer 4

but in the inner space, it works as annihilator

Answer 5

So these vector spaces are just sets of vectors?

Answer 6

No, they are subspace themselves

Answer 7

\(\cap\) is intersection

Answer 8

What does + do for vector space?

Answer 9

I think direct sum

Answer 10

How do you sum vector space?

Answer 11

why not? We have theorem showing it
Let say {m1......m_m, n1,...........,n_n} is basis of V and M +N = V (you know what I mean, right?)

Answer 12

Union?

Answer 13

direct sum, how ever, if M union N = {0} , then M +N = V no need to use direct sum here

Answer 14

I have the definition of \(\perp\) set
S \(\subset \)V , \(S^\perp ={\vec u \in V:(\vec u,\vec v)=0 for~~ all~~ \vec v \in S}\)

Answer 15

\[
\forall m,n\quad \exists v\quad ((m+n),v)= 0 \iff \exists u,w \quad (u,m) = 0\land (w,n)
\]

Answer 16

\[
\forall m,n\quad \forall v\in V\quad ((m+n),v)= 0 \iff \exists u,w\in V \quad (u,m) = 0\land (w,n)
\]

Answer 17

@wio, I don't get the last term from (u,m) =0 \(\land (w,n)\)

Answer 18

Both are just \((\_,\_) = 0\), since they are perpendicular

Answer 19

The only problem is that I'm confused about the definition of + and \(\cap\) in this case.

Answer 20

I need read more to understand the stuff :),

Answer 21

This is my proof, need check, please
from the left hand side
\((M+N)^\perp = set~~of~~{\vec u: (\vec u, \vec m+\vec n) =0~~ \forall \vec m\in M ~~and~~\vec n\in N}\)
now open the right hand side of the above \((\vec u, \vec m+\vec n) = (\vec u,\vec m)+(\vec u, \vec n)=0\)
since M \(\neq \)N, so that the sum = 0 iff each element =0
\((\vec u, \vec m)=0 ~~and~~(\vec u, \vec n)=0\)
moreover, {\(\vec u: (\vec u, \vec m)=0\)} = \(M^\perp\)
and {\(\vec u: (\vec u, \vec n)=0\)} = \(N^\perp\)
so, \((M+N)^\perp = M^\perp \cap N^\perp\)\(\color{red}{\text{it's so weak, need correct, please}}\)

Answer 22

@Miracrown

Answer 23

@KingGeorge

Answer 24

Your argument looks fine. The only issue I see is that it only proves that\[(M+N)^\perp \subseteq M^\perp \cap N^\perp\]and you want both directions.

Answer 25

Nevermind. I missed the "iff" you had in there. It looks fine to me.

Answer 26

but I don't satisfy with my argument:( how "and" becomes \(\cap\) ?
if it is \(M^\perp + N^\perp\), it make more sense than it turn to " \(\cap\) "

Answer 27

Well, by definition, if a vector \(\vec{v}\) is in \(M^{\perp}\) and \(N^\perp\), then it must be in \(M^\perp\cap N^\perp\). That's just the definition of \(\cap\).

Answer 28

Thank you. :)