I just solved a similar question but with 'OR' instead of 'AND'.Dick said:C[-1,1] doesn't have anything to with polynomials. You have to show if f(x) and g(x) satisfy your condition then so does f(x)+g(x) and c*f(x). Forget the polynomials.
Dick said:You can use a polynomial to provide give a counterexample in the OR case. Because it's false and the polynomials are contained in C[-1,1]. You can't prove the case of AND just using polynomials because it is true.
Dick said:Just show f(x) and g(x) vanish at both endpoints that f(x)+g(x) and c*f(x) also have that property.
Roni1985 said:Ohh, I see...
But, what are my f(x) and g(x) ?
If I can't use the polynomials, what can I use ?
are the simply the 0 functions ?
f(x)=0=g(x)
?
OMG, you are so right ...Dick said:f(x) and g(x) are just continuous functions on [-1,1]. All you know about them is that f(-1)=f(1)=0 and g(-1)=g(1)=0, since they are in the set that you are supposed to prove is a subspace. That's all you need to know. Isn't that enough?
A subspace in linear algebra is a subset of a vector space that contains all the necessary properties to also be considered a vector space. This means that it must contain the zero vector, be closed under vector addition and scalar multiplication, and must contain all possible linear combinations of its vectors.
To determine if a set of functions in C[-1,1] is a subspace, you must check if it satisfies the three main properties of a subspace: closure under vector addition, closure under scalar multiplication, and containing the zero vector. This can be done by checking if the set of functions contains the zero function, if it is closed under addition and scalar multiplication, and if all linear combinations of its vectors are also in the set.
Yes, a subspace in C[-1,1] can contain an infinite number of functions. This is because the set of all continuous functions in the interval [-1,1] is an infinite-dimensional vector space, so any subspace of it can also be infinite-dimensional.
Subspaces in C[-1,1] are related to linear independence in that if a set of functions in C[-1,1] is linearly independent, then it forms a basis for a subspace in C[-1,1]. This means that all other functions in the subspace can be expressed as linear combinations of the basis functions.
Yes, a subspace in C[-1,1] can contain functions of different degrees. This is because the degree of a function does not affect its properties as a vector, so as long as the set of functions satisfies the properties of a subspace, it can contain functions of different degrees.