Proving the Properties of Subspace U⊥ in Rn | Help with Subspace Concepts

[icystrike]

1. Let U be a subspace of Rn and let
U⊥ = {w ∈ Rn : w is orthogonal to U} .
Prove that
(i) U⊥ is a subspace of Rn,
(ii) dimU + dimU⊥ = n.


Attempt.


i)
U. ( U⊥)T=0
If U⊥ does not passes the origin , the above equation cannot be satisfied.
Therefore U⊥ passes the origin.
U.( U⊥+ U⊥)T=U. ( U⊥)T+U.( U⊥)T=0+0=0
U.(k U⊥)T=k[U. (U⊥)T]=k.0=0

Therefore U⊥ is a subspace in Rn

ii) let U have rank r.
U⊥ is the nullspace of the U transpose.
and if U is a matrix of mxn , UT is nxm.
Therefore , the sum of dim of the two matrices is exactly n.

 

[HallsofIvy]

I don't understand your notation. You use "T" which I guess is "transpose" and talk about U being a matrix. U is given as a subspace, not a matrix.

 

[icystrike]

I don't understand your notation. You use "T" which I guess is "transpose" and talk about U being a matrix. U is given as a subspace, not a matrix.

yup you are right.. it is transpose and subspace..

 

[adityab88]

1. Let U be a subspace of Rn and let
U⊥ = {w ∈ Rn : w is orthogonal to U} .
Prove that
(i) U⊥ is a subspace of Rn,
(ii) dimU + dimU⊥ = n.


Attempt.


i)
U. ( U⊥)T=0
If U⊥ does not passes the origin , the above equation cannot be satisfied.
Therefore U⊥ passes the origin.
U.( U⊥+ U⊥)T=U. ( U⊥)T+U.( U⊥)T=0+0=0
U.(k U⊥)T=k[U. (U⊥)T]=k.0=0

Therefore U⊥ is a subspace in Rn

ii) let U have rank r.
U⊥ is the nullspace of the U transpose.
and if U is a matrix of mxn , UT is nxm.
Therefore , the sum of dim of the two matrices is exactly n.

Let u,v be in U⊥. Let k be a constant in R.
Now <u,w>=0 for all w in U and <v,w>=0 for all w in U.
Thus by linearity of inner product we have <u+kv,w>=0 for all w in U. Thus, u+kv is also in U⊥ for all u,v,k. Thus U⊥ is a subspace.

For the second part, DIY but here are some hints:
- We know that U $$\cap$$ U⊥ = {0}. Easy to check (if a vector is in U and perpendicular to U then it has to be the zero vector).
- Take x in Rn. Prove that x = u+u' for u in U and u' in U⊥.
- This is unique representation. If x = u+u' = v+v' with u,v in U and u',v' in U⊥ then 0 = (u-v)+(u'-v') implies u-v = v'-u' implies u-v = v'-u' = 0 since U $$\cap$$ U⊥ = {0}. Thus u=v and u'=v'.
- Thus projection map $$\pi$$: Rn -->> U defined by $$\pi$$(x)=u where x = u+u' with u in U and u' in U⊥.
- Use Isomorphism theorem corollary (that dim(range)+dim(kernel)=dim(Rn)=n)

 

[icystrike]

Let u,v be in U⊥. Let k be a constant in R.
Now <u,w>=0 for all w in U and <v,w>=0 for all w in U.
Thus by linearity of inner product we have <u+kv,w>=0 for all w in U. Thus, u+kv is also in U⊥ for all u,v,k. Thus U⊥ is a subspace.

For the second part, DIY but here are some hints:
- We know that U $$\cap$$ U⊥ = {0}. Easy to check (if a vector is in U and perpendicular to U then it has to be the zero vector).
- Take x in Rn. Prove that x = u+u' for u in U and u' in U⊥.
- This is unique representation. If x = u+u' = v+v' with u,v in U and u',v' in U⊥ then 0 = (u-v)+(u'-v') implies u-v = v'-u' implies u-v = v'-u' = 0 since U $$\cap$$ U⊥ = {0}. Thus u=v and u'=v'.
- Thus projection map $$\pi$$: Rn -->> U defined by $$\pi$$(x)=u where x = u+u' with u in U and u' in U⊥.
- Use Isomorphism theorem corollary (that dim(range)+dim(kernel)=dim(Rn)=n)

Hmm.. i think for part 2 i can explain by stating that dim(U+U⊥)=dim(U)+dim(U⊥)+dim(U$$\cap$$ U⊥)
Rn=dim(U)+dim(U⊥)
since dim(U$$\cap$$ U⊥) ought to be atleast {0} to satisfy the condition of subspace , and all the more they are orthogonal to one another , the can't be parallel , thus they have to be {0} itself.
Hence proving part 2.