# Finding a Basis for the Kernel of a Differential Operator

• simpledude
In summary: Ok so this is a way of solving the equation for x, but you still need to solve it for a and b.In summary, The solution to the equation for x is when a and b are both 0. The basis for Ker T is {e^-2x, e^x}.
simpledude

## Homework Statement

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

solve for u and show the solutions are linearly independent

Last edited:
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!

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 )

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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.

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.

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).

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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}

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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.

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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 let's 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

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.

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simpledude said:
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.

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

Thank you sir!

## What is a kernel in linear algebra?

A kernel, also known as null space, is the set of all vectors that map to the zero vector when multiplied by a linear transformation. In other words, it is the set of all inputs that result in an output of zero.

## Why is finding a basis of a kernel important?

Finding a basis of a kernel is important because it allows us to understand the behavior of a linear transformation and its relationship to the input vectors. It also helps us to solve systems of linear equations and find solutions to problems in a more efficient way.

## How do you find a basis of a kernel?

To find a basis of a kernel, we need to solve the homogeneous system of linear equations associated with the linear transformation. This can be done by setting up an augmented matrix and using row reduction techniques to find the pivot columns, which will form the basis of the kernel.

## Can a kernel have more than one basis?

Yes, a kernel can have more than one basis. This is because there could be multiple ways to represent the same set of vectors as a linear combination. However, all bases of a kernel will have the same number of vectors, known as the dimension of the kernel.

## What is the relationship between the kernel and the image of a linear transformation?

The kernel and the image of a linear transformation are complementary subspaces. This means that the dimension of the kernel plus the dimension of the image equals the dimension of the input vector space. In other words, the kernel and the image span the entire input vector space.

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