# Orthogonal theoretical question

#### transgalactic

V is a space of inner muliplication.
W1 and W2 are two subspaces of V, so dimW1<dimW2
prove that there is a vector
$$0\neq v\epsilon W_2$$
which is orthogonal to W1
??

Related Calculus and Beyond Homework Help News on Phys.org

#### transgalactic

i got this solution:
dim W1=k
dim W2=n

$$n-k=W_2^\perp$$
i cant see why its correcti dont know that all the vectors of W1 exist in W2 too
??

Last edited:

#### yyat

Since dim(W1)<dim(W2) you can find a nonzero vector v in W2 which is not an element of W1. Use the orthogonal projection onto W1 to construct, from v, a vector which is orthogonal to W1.

#### transgalactic

its a theoretical question i cant construct projections
the only way is if this "
dim W1=k
dim W2=n

$$n-k=W_2^\perp$$
"

if all the vectors of W1 exist in W2
which wasnt gisnt given that
??

#### yyat

its a theoretical question i cant construct projections
What I am trying to tell you is this: If V is an inner product space and W is subspace, then any vector can be written as a sum v=w+n where w is in W and n is in the orthogonal complement of W. In fact, w is the projection of v to W. So, assuming that v is not an element of W, then n will be nonzero and orthogonal to W.
Usings these facts you will be able to solve the problem.

the only way is if this "
dim W1=k
dim W2=n

$$n-k=W_2^\perp$$
"

if all the vectors of W1 exist in W2
which wasnt gisnt given that
??
You do not need W1 to be a subspace of W2.

#### transgalactic

whats
"V is an inner product space "
??

Last edited:

#### transgalactic

i am trying to imagine what you are telling
but its so abstract
??

#### transgalactic

suppose we have a four dimentional space
W={(1,0,0,0),(0,1,0,0)}
W complement is ={(0,0,1,0),(0,0,0,1)}
w is subspace of W
n is subspace of W complement
v could be written as w+n
but for that W and W complement must be independant to one another

we havent been told that
we have been told that
dim W2>dim W1

so there could a case were W2 has all the vectors of W1 and there is another subspace

so not every vector could be written as a manipulation of W2 and W1
??

#### yyat

whats
"V is an inner product space "
??
A http://en.wikipedia.org/wiki/Inner_product_space" [Broken], I am assuming this is what you mean by "a space of inner multiplication".

suppose we have a four dimentional space
W={(1,0,0,0),(0,1,0,0)}
W complement is ={(0,0,1,0),(0,0,0,1)}
w is subspace of W
n is subspace of W complement
v could be written as w+n
but for that W and W complement must be independant to one another

we havent been told that
we have been told that
dim W2>dim W1

so there could a case were W2 has all the vectors of W1 and there is another subspace

so not every vector could be written as a manipulation of W2 and W1
??
I am not sure what you mean here, I will try to explain it differently.

What you are given: $$W_1$$, $$W_2$$ are subspaces of $$V$$ with $$\dim(W_1)<\dim(W_2)$$.
Let's put $$n=\dim(V)$$, $$n_1=\dim(W_1)$$, $$n_2=\dim(W_2)$$.

So $$\dim(W_1^{\perp}\cap W_2)\ge (n-n_1)+n_2-n\ge 1$$.

But this means that there are nonzero vectors in $$W_2$$ which are orthogonal to $$W_1$$.

I hope you can fill in the details, specifically, why the inequality I wrote is true.

Last edited by a moderator:

#### transgalactic

i cant understanf how did you constract this equation
$$\dim(W_1^{\perp}\cap W_2)\ge (n-n_1)+n_2-n\ge 1$$
if$$(W_1^{\perp}=n-n_1$$ and $$n_2=dim(W2)$$
so the intersect is subtraction of the two groups
it should be
$$W_1^{\perp}\cap W_2=n-n_1-n_2$$

and why
you add =>1
??

#### yyat

i cant understanf how did you constract this equation
$$\dim(W_1^{\perp}\cap W_2)\ge (n-n_1)+n_2-n\ge 1$$
I am using the following formula: If A,B are subspaces of V then

$$\dim(A+B)+\dim(A\cap B)=\dim(A)+\dim(B)$$

and

$$\dim(A+B)\le\dim(V)$$

if$$(W_1^{\perp}=n-n_1$$ and $$n_2=dim(W2)$$
so the intersect is subtraction of the two groups
it should be
$$W_1^{\perp}\cap W_2=n-n_1-n_2$$

and why
you add =>1
??
If dim(V)>0 then V contains nonzero vectors.

A general comment: A message board is not the best way to learn linear algebra. You need someone you can talk to face to face, a teacher or other students. If you are stuck on a specific problem, then the people here can give you hints, but this only works if you already know most of what you should know to solve a particular exercise.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving