Vector Space & Vector Subspaces

In summary, the problem is that the set of solutions to the differential equation is a vector subspace of the vector space of all continuours functions, but the zero vector needs to be included. The solution set is a subset of V, and the two functions that comprise it are its basis.
  • #1
rad0786
188
0
Hello...

I've been doing some home work on Vector Spaces and Vector Subspaces and I need help solving a problem... Can somebody please help me?

Consider the differential equation f'' + 5f' + 6f' = 0 Show that the set of all solutions of this equation is a vector subspace of the vector space of all continuours funtions with the usual operations


How I went about this question is considering the 3 for U to be a vector subspace of V
1. The zero vector has to be in U
2. if r and s are in U, then r+s lies in U
3. if r lies in U, then kr lies in U for all k in R

I know that the zero vector is in U since f'' + 5f' + 6f' = 0 (IT EQUALS 0) however, I don't know how to show the other 2 conitions. (conditions 2 and 3.)

Can somebody PLEASE help me.

Thank you
 
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  • #2
rad0786 said:
Consider the differential equation f'' + 5f' + 6f' = 0

Should that last term on the LHS be primed? I'm thinking not, otherwise you would have just combined the the last two terms on the left.

however, I don't know how to show the other 2 conitions. (conditions 2 and 3.)

Solve the differential equation. The general solution will be a linear combination of 2 functions. If your solution set is a subspace, then those two functions will be its basis.
 
  • #3
Sorry, yeah, the last one shouldn't have been primed ---- > f'' + 5f' + 6f = 0

Okay i will try that, thanks!
 
  • #4
I don't know how to show the other 2 conitions. (conditions 2 and 3.)

They follow immediately from the fact that the derivative is linear, i.e. (g(x) + h(x))' = g'(x) + h'(x) and (k * g(x))' = k * g'(x).
 
  • #5
Muzza, I still don't get it :(

so if you take the derivative of f'' + 5f' + 6f = 0 then it becomes linear and that just proves that condition?

I mean, I was thinking of what Tom Mattson said above, that too was complicated.

So i guess the fact that the derivitave is linear automatically solves condition 2 and 3?
 
  • #6
Suppose h and g are in the set of solutions of that differential equation.

Among other things, you wish to prove that h + g is also in that set. So what do you do? You simply verify that (h(x) + g(x))'' + 5(h(x) + g(x))' + 6(h(x) + g(x)) is equal to 0 (which follows immediately from the linearity of the derivative).
 
  • #7
ohhh I see how it works... what you did was substitute (h(x) + g(x)) into each of the f. but i still don't see how you get something linear out of this and how it shows condition 2?
 
  • #8
You don't "get something linear" out of it, it just is linear. That is, the derivative operator satisfies:

[tex]\frac{d}{dx}(y+z)=\frac{dy}{dx}+\frac{dz}{dx}[/tex]
[tex]\frac{d}{dx}(ky)=k\frac{dy}{dx}[/tex],

for all differentiable functions [itex]y[/itex] and [itex]z[/itex].
 
  • #9
Hello again. So i did some research and thought about whatyou said. I came up with this... can u please tell me if I am on the right path.

Conditon 1: Simply substitue O into f

(0) + 5(0) +6(0) = 0

Condition 2: Show that h(x) and g(x) is in the subspace

(h(x) + g(x))'' + 5(h(x) + g(x))' + 6(h(x) + g(x))'' = 0

Condition 3: Show the closure of scalar multiplication

r * (f'' + 6f' +5f) = r * 0
(f'' + 6f' +5f) = 0


Can somebody please tell me if that works?

Thanks
 
  • #10
Well, it will if you finish the calculations! You want to use the fact that f and g satisfy the equation themselves to show that f+ g and rf satisfy the equations.

On the second one, in my opinion, you need to start "back a step"- start with
(rf)"+ 6(rf)'+ (rf) and see what that gives you. And don't start by saying the are equal to 0- that's what you want to show.
 
  • #11
Dear rad, let me show u how to show condition 2;
suppose f and g are in the set. ie f'' + 5f' +6f = 0 and g'' + 5g' + 6g=0

Now, u want to show (f+g) is in the set. ie u want to show
(f+g)''+5(f+g)'+6(f+g)=0


Start on left hand side:
(f+g)''+5(f+g)'+6(f+g)
=(f'' +5f' + 6f)+(g'' +5g' +6g)
=0+0
=0
=right hand side.

You're done
 
  • #12
This is all starting to become more clear!

So i suppose that condtion 3 would be ike this:

Let r be any scalar in the set of Real numbers... and we want to show that r multiplied by f'' + 6'' + 5'' gives an answer of 0


Left Hand Side
(rf)"+ 6(rf)'+ 5(rf)
= r x (f'' + 6f' + 5f)
= r x 0
= 0 = Right Hand Side

Tzar... the on thing that got me on condtion 2, and still gets me... is how you went from (f+g)''+5(f+g)'+6(f+g) to (f'' +5f' + 6f)+(g'' +5g' +6g)
What's confusing me is the primes

Thanks.
 
  • #13
rad0786 said:
This is all starting to become more clear!

So i suppose that condtion 3 would be ike this:

Let r be any scalar in the set of Real numbers... and we want to show that r multiplied by f'' + 6'' + 5'' gives an answer of 0


Left Hand Side
(rf)"+ 6(rf)'+ 5(rf)
= r x (f'' + 6f' + 5f)
= r x 0
= 0 = Right Hand Side

Thats right!


rad0786 said:
Tzar... the on thing that got me on condtion 2, and still gets me... is how you went from (f+g)''+5(f+g)'+6(f+g) to (f'' +5f' + 6f)+(g'' +5g' +6g)
What's confusing me is the primes

Thats because differntiation is linear. Ie (f+g)' = f' +g' and (rf)' = rf' if r is a constant
 
  • #14
Thanks so much! I caught on quickly after doing some more question out of the textbook. It was complicated because I am not used to seeing differentation in linear algebra, in fact, we didnt even learn how to solve equations like that.

In the mean time, another question has got me, which is related to subspaces.

" Suppose U and W are vector subspaces of a real vector space V. The union U u W of U and W is the set of all vectors which lie either in U or in W. Suppose we know that U u W is a subspace of V. Whow that either U c W or W c U. "

This is how I tired this question ... Suppose that U was not in W (or W not in U) Then a vectors in only in U cannot be in W since the vectors are in different subspaces. However, a vector that satisfies the condtions of a subspace of U and W, (the intersection of U and W) then it is part of the U u W. That means that U has to be in W or W has to be in U.

I know that sounds very confusing, but I have no idea how to show that algebraically.

Am i on the right track?
 
  • #15
You are definitely on the right track but you have to be more explicit about the statement "a vector that satisfies the conditions of a subspace of U and W, (the intersection of U and W) then it is part of the U u W" - you need to give the details there.
 

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and two operations, addition and scalar multiplication, that satisfy certain properties. These properties include closure, associativity, commutativity, and the existence of an additive identity and inverse, among others. Vector spaces are used in many areas of mathematics, including linear algebra and functional analysis.

2. What is a basis for a vector space?

A basis for a vector space is a set of vectors that are linearly independent and span the entire vector space. This means that any vector in the vector space can be written as a unique linear combination of the basis vectors. The number of vectors in a basis is called the dimension of the vector space.

3. How do you determine if a set of vectors is a basis for a vector space?

To determine if a set of vectors is a basis for a vector space, you can use the following steps:

  • Check that the vectors are linearly independent by setting up a linear combination and solving for the coefficients. If the only solution is the trivial solution (all coefficients are zero), then the vectors are linearly independent.
  • Check that the vectors span the entire vector space by ensuring that any vector in the space can be written as a linear combination of the basis vectors.
  • If both of these conditions are met, then the set of vectors is a basis for the vector space.

4. What is a vector subspace?

A vector subspace is a subset of a vector space that satisfies the properties of a vector space. This means that the subset must be closed under addition and scalar multiplication, and must also contain the additive identity and inverse. A vector subspace can also be thought of as a smaller vector space within a larger vector space.

5. How do you determine if a subset is a vector subspace?

To determine if a subset is a vector subspace, you can use the following steps:

  • Check that the subset is non-empty and contains the zero vector.
  • Check that the subset is closed under addition, meaning that if you add any two vectors in the subset, the result is also in the subset.
  • Check that the subset is closed under scalar multiplication, meaning that if you multiply any vector in the subset by a scalar, the result is also in the subset.
  • If all of these conditions are met, then the subset is a vector subspace.

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