Vector Space & Vector Subspaces

[rad0786]

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

 

[quantumdude]

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.

 

[rad0786]

Sorry, yeah, the last one shouldn't have been primed ---- > f'' + 5f' + 6f = 0

Okay i will try that, thanks!

 

[Muzza]

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).

 

[rad0786]

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?

 

[Muzza]

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).

 

[rad0786]

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?

 

[quantumdude]

You don't "get something linear" out of it, it just is linear. That is, the derivative operator satisfies:

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

for all differentiable functions y and z.

 

[rad0786]

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

 

[HallsofIvy]

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.

 

[Tzar]

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

 

[rad0786]

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.

 

[Tzar]

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!

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

 

[rad0786]

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?

 

[David]

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.