# Infinite dimensional counterexample

1. Dec 11, 2008

### Carl140

1. The problem statement, all variables and given/known data

Let V be a finite dimensional vector space and let W be a subspace of V.
1. Then V is the direct sum of W and W' where W' denotes the orthogonal complement of W.
2. Also, (W')' = W, i.e the orthogonal complement of the orthgonal complement of W is
again W.

My question is, what happens if we drop the condition that V is finite dimensional, would
the results would be still valid? what happens with condition 1 and 2??

2. Dec 11, 2008

### HallsofIvy

Staff Emeritus
Part of the problem here is that "orthogonality" depends on the specific inner product and there is no "natural" inner product on infinite dimensional vector spaces. Are you assuming a specific inner product and, if so, what?

3. Dec 11, 2008

### Carl140

Sorry, I accidentally clicked on report.

I read somewhere in the web that l^2(N) is such counterexample where l^2(N) is the set of all sequences of real numbers (x_1, x_2,...) such that
sum( x_i^2 , i=1 to infinity) < infinity.

But I don't know which subspace of l^2(N) should I consider to find where the properties fail.

4. Dec 11, 2008

### Dick

I'll give you a simpler example. Let V be the space of all continuous functions on [0,1]. Define the inner product <f,g> to be the integral of f*g over [0,1]. Let W be the subspace of all functions such that f(0)=0. The only element of W' is f=0. V obviously isn't equal to the direct sum of W and W'. They don't even span V.

5. Dec 12, 2008

### Dick

In the case of I^2(N), take W to be the subspace of all sequences with only a finite number of elements nonzero. Both of these examples have a common elements. You have a proper subspace W that is dense in V.