# Finding a basis of a Kernel

1. Mar 5, 2008

### simpledude

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

Let V = C(R,R) be the vector space of all functions f : R −> R that have continuous
derivatives of all orders. We consider the mapping T : V −> V defined for all u belonging to V , by T(u(x)) = u''(x) + u'(x) − 2u(x). (Where u' is first derivative, u'' second derivative)

Question: Determine Ker{T} and find a bsis of Ker{T}

2. The attempt at a solution

So as far as I understand the Ker{T} is the set of elements in T that maps into
the zero vector, i.e. Ker{T} is when T(u(x))=0 (is this correct?)

So T(u(x)) = 0 only when u''(x) + u'(x) = 2u(x). So would the Kernel of T be all functions
u belonging to V such that u''(x) + u'(x) = 2u(x)

But in this case how can I compute the basis?

---
Thank you very much.
Best Regards,
SimpleDude

2. Mar 5, 2008

### ircdan

solve for u and show the solutions are linearly independent

Last edited: Mar 5, 2008
3. Mar 5, 2008

### simpledude

Ok, so u(x) = [u''(x) - u'(x)] / 2

But that's only one solution, how can I get anoter one from the same equation?
(because if I use the same equaton, they both will be linearly dependant).

Any help will be appreciated -- thanks!

4. Mar 5, 2008

5. Mar 5, 2008

### simpledude

Ok I understand, and as far as the initial conditions go, if T is a linear operator
is it ok to say that u(0)=0 and u'(0)=0 ?

Because in that case everythign is 0 (both constants are 0, hence u(x)=0 )

Last edited: Mar 5, 2008
6. Mar 5, 2008

### ircdan

no you won't have initial conditions, so if you solved it(i'm assuming you did), you should get that u is of the form,

u(x) = ae^(x) + be^(-2x) for some a, b in R, in particular, this means that

u belongs to span({e^(x), e^(-2x)} = B(I just called it B for notation), so now you've shown
that kerT is contained in spanB, and the other inclusion is obvious(you should check it to be sure), so kerT = spanB

Now just show it's independent, this might be trickier for you, so i'll get you started. We'll just use the definition of linear independence.

So suppose ae^x + be^(-2x) = 0 for some scalars a, b, in R.

We need to show a = b = 0.

hint: this equation must hold for ALL values of x, so try some different values and try to force a = b = 0.

7. Mar 5, 2008

### simpledude

Ahhh :)

Can I create a matrix of coefficients of
u(x) and u'(x), namely:

u(x) = ae^-2x + be^x
u'(x) = -2ae^-2x + be^x

And solve AX=0 via gauss?

Doing so I get a matrix of the form
1 1
0 1

Hence the are linearly independant, furhermore (1,0) and (1,1) form a basis of Ker T.

8. Mar 5, 2008

### ircdan

The B I wrote above will be your basis, remember B is a subset of V which is a space of functions, (1, 0) and (1, 1) are elements of R^2, they certainly aren't functions!

I gave you a hint, it's a good one, try it!:)

Edit: Another way is to compute the wronskian of e^(-2x) and e^x and notice it's nonzero, but imo it's better to do it the way I suggested. What I suggested is a good trick to keep around because it's useful in other places(like when working with dual spaces, etc).

Last edited: Mar 5, 2008
9. Mar 5, 2008

### simpledude

Ok, that's not that hard.

Since: e^x != e^-2x for any values of x, except 0, and e^0 = 1.
The only way to satisfy the equation ae^x + be^x = 0
is if th coefficients a,b=0 i.e. a=b=0

Hence it is linearly independat and the basis of Ker T is indeed {e^-2x, e^x}

Last edited: Mar 5, 2008
10. Mar 5, 2008

### ircdan

that's not enough, the way you wrote it isn't really correct, but the idea is correct!

We have ae^x + be^(-2x) = 0, we need to show a = b = 0. This equation holds for all values of x. So plug some values of x in.

Set x = 0. Then we get a + b = 0, so b = -a. Now go back to the original equation,

ae^x + be^(-2x) = ae^x - ae^(-2x) = a(e^x - e^(-2x)) = 0. Then a = 0, otherwise e^x - e^(-2x) = 0, that is e^x = e^(-2x) and this holds for all x in R, a contradiction(for example setting x = 1 shows it's clearly false).

Therefore b = -a = -0 = 0, so a = b = 0 and we have independence, and B is a basis as needed.

EDIT: I edited my post to give you more since I noticed you were really close.

I hope this helps, it's a good trick to keep around.

Last edited: Mar 5, 2008
11. Mar 5, 2008

### simpledude

Ok so more abstratly,

Let's take an arbitrarily large x, x --> infinity
The equation e^x will go to infinity
The equation e^-2x will go to zero.

So a(infinity) + b(0) = 0 is only satisfied if a=0

Now lets take x --> -(infinity)
Th equation e^x will go to zero.
The equation e^-2x will go to infinity.

So a(0) + b(infinitiy)=0 is only satisfied if b=0

Hence a=b=0

12. Mar 5, 2008

### simpledude

hehe thanks!
However, is my approach correct?
Can infinity be used?

Also Dan, if you don't mind me asking a more abstract question.
Let's say I wanted to show that T was a linear operator.
Can I independantly show each part of the operator? I mean can I show
u''(x) is linear, then u'(x) is linear, and -2u(x) is linear.

Then since they are all linear, a linear combination u''(x)+u'(x)-2u(x)
is also linear, hence T is a linear operator.

Last edited: Mar 5, 2008
13. Mar 5, 2008

### ircdan

hmm no not really to the infinity thing, for practice you could show that {e^(bx), e^(cx)} is linearly independent if c != b. (using the same method I described)

As for showing T is linear, yes, exactly what you said! It's actually easier, from calculus you know that differentiation is a linear operator, ie, d/dx(u + v) = du/dx + dv/dx, d/dx(cu) = c*du/dx, and the same when you take the second derivative because the composition of linear operators is linear, so we had

T(u) = u'' + u' - 2u.

So from calculus, it's immediate that
T(u + v) = (u + v)'' + (u + v)' - 2(u + v)
= u'' + v'' + u' + v' -2u - 2v
= u'' + u' - 2u + v'' + v' - 2v
= T(u) + T(v)

and similarly T(cu) = cT(u)

where we used the fact that differentiation is a linear operator

14. Mar 5, 2008

### simpledude

Thank you sir!