Linear Algebra. Proving differentiable functions are a vector space.

In summary: Therefore, the set of all differentiable functions on ℝ that satisfy f'+2f=0 is a vector space.In summary, the set of all differentiable functions on (-infinity, +infinity) that satisfy f′ + 2f = 0 is a vector space, denoted by U. To prove that U is a subspace of the larger vector space V, it must be shown that U contains the zero function and is closed under addition and scalar multiplication. This can be done by proving that (f+g) satisfies the given equation, making f and g variables in the vector space. This helps clarify the concept and provides a better understanding for solving similar problems.
  • #1
datran
14
0
Question: Show the set of all differentiable functions on (-infinity, +infinity) that satisfy f′ + 2f = 0 is a vector space.

I started the problem by assuming that f and g are both differentiable functions that satisfy this vector space.

Then I ran through the ten axioms of addition and scalar multiplication and proving that each one works.

I feel like that does not answer the question though since why would I need the equation f' + 2f = 0?

How does that equation come into play?

Thanks for any help provided.
 
Physics news on Phys.org
  • #2
welcome to pf!

hi datran! welcome to pf! :smile:
datran said:
I feel like that does not answer the question though since why would I need the equation f' + 2f = 0?

How does that equation come into play?

you have to prove eg that (f+g) satisfies that equation :wink:

(yes, i know it's obvious … but you still have to prove it!)
 
  • #3
Oh!

So I would do (f+g) = (f+g)' + 2(f+g) = 0

and same thing over and over for the 10 axioms.

So really f and g are like variables?

Thank you so much! That actually made many more problems clearer!
 
  • #4
You can start by proving that the set of all differentiable functions from ℝ to ℝ with the standard definitions of addition and scalar multiplication is a vector space. (Looks like you've done that already). Denote this space by V. Define U={f in V|f'+2f=0}. U is by definition a subset of V. If you prove that U contains the 0 function and is closed under addition and scalar multiplication, you can conclude that U is a subspace of V.
 
  • #5


In order to prove that the set of all differentiable functions on (-infinity, +infinity) that satisfy f′ + 2f = 0 is a vector space, we need to show that it satisfies the ten axioms of a vector space: closure under addition, closure under scalar multiplication, commutativity of addition, associativity of addition, existence of an additive identity, existence of an additive inverse, distributivity of scalar multiplication over addition, distributivity of scalar multiplication over scalar addition, associativity of scalar multiplication, and existence of a multiplicative identity.

The equation f' + 2f = 0 is important because it is the defining property of the set of functions we are considering. It is the condition that all functions in this set must satisfy. This condition ensures that the set is closed under addition and scalar multiplication, as well as satisfying the other axioms.

To show closure under addition, we need to show that if f and g are both differentiable functions that satisfy f' + 2f = 0, then their sum f+g must also satisfy f' + 2(f+g) = 0. Similarly, to show closure under scalar multiplication, we need to show that if f is a differentiable function that satisfies f' + 2f = 0, then kf also satisfies f' + 2(kf) = 0 for any scalar k.

By using the properties of differentiation and the fact that f' + 2f = 0, we can easily show that the sum and scalar multiple of two functions that satisfy this equation will also satisfy it. This shows that the set is closed under addition and scalar multiplication.

The other axioms can also be proven using the equation f' + 2f = 0. For example, to show commutativity of addition, we can use the fact that differentiation is a linear operator and thus the order of addition does not matter. Similarly, the existence of an additive identity can be proven by showing that the function f(x) = 0 satisfies the equation f' + 2f = 0.

In conclusion, the equation f' + 2f = 0 is crucial in proving that the set of all differentiable functions on (-infinity, +infinity) that satisfy this equation is a vector space. It is the condition that ensures the set satisfies all ten axioms of a vector space.
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with vector spaces and linear mappings between them. It involves the study of linear equations, matrices, and their properties.

2. What is a vector space?

A vector space is a collection of objects called vectors that can be added together and multiplied by numbers, called scalars. It is a fundamental concept in linear algebra and is used to represent real-world quantities such as forces, velocities, and coordinates.

3. How do you prove that a set of differentiable functions is a vector space?

To prove that a set of differentiable functions is a vector space, we must show that it satisfies the following properties: closure under addition and scalar multiplication, associativity, commutativity, existence of an identity element, and existence of inverse elements. Then, we can use the definition of a vector space to formally prove that the set of differentiable functions satisfies all of these properties.

4. Can you give an example of a differentiable function that is not a vector space?

Yes, the set of all polynomials of degree less than or equal to 3 is not a vector space because it does not satisfy the closure property under scalar multiplication. If we multiply a polynomial by a scalar, the resulting function may have a degree higher than 3, which is not in the original set.

5. How is linear algebra used in real life?

Linear algebra has a wide range of applications in various fields, including physics, engineering, computer science, and economics. It is used to solve systems of equations, analyze data and patterns, and develop algorithms for machine learning and artificial intelligence. It also has practical applications in fields such as graphics and image processing, cryptography, and network analysis.

Similar threads

  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
770
  • Linear and Abstract Algebra
Replies
4
Views
766
  • Linear and Abstract Algebra
Replies
3
Views
963
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
2K
Replies
5
Views
484
  • Linear and Abstract Algebra
Replies
10
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
755
Replies
2
Views
935
Back
Top