Finding kernel and range for polynomials transformation

[charmmy]

I have troubles arriving at the solution to this question:
Consider the transformation T: P3-->P3 given by:
T(f)=(1-x^2)f '' - 2xf '

Determine the bases for its range and kernel and nullity and rank

Can anyone explain how should i go about finding the bases for its kernel and range??

i get 0 for the nullity, which I assume is wrong??

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Yes, of course, 0 nullity is wrong. A linear transformation has trivial kernel if and only if it is invertible. And differentiation is not invertible.

I am assuming that P3 means the vector space of polynomials of degree 3 or less, which has dimension 4. Some texts use P3 to mean the 3 dimensional vector space of polynomials of degree 2 or less. If that is the case, the same argument works and is simpler.

The kernel consists of all vector in P3, that is all polynomials of the form $p(x)= a+ bx+ cx^2+ dx^3$, such that $T(p)= (1- x^2)p''- 2xp'=$$(1- x^2)(2c+ 3dx)- 2x(b+ 2cx+ 3dx^2)=$$2c+ 3dx- 2cx^2- 3dx^3- 2bx- 4cx^2- 6dx^3$$= -9dx^3- 6cx^2+ (3d- 2b)x+ 2c= 0$, for all x. That means we must have -9d= 0 so d= 0, -6c= 0 so c= 0, 3d- 2b= -2b= 0 so b= 0, but there is NO condition on a. Any polynomial of the form $tex= a+ 0x+ 0x^2+ 0x^3= a$ is in the kernel of T. The nullity is 1 and a basis for the kernel is the single constant polynomial {1}.

Since the nullity has dimension 1 and P3 has dimension 4, the range must have dimension 4- 1= 3.

Again, writing $p(x)= a+ bx+ cx^2+ dx^3$ we have, as before, $T(p)= -9dx^3- 6cx^2+ (3d- 2b)x+ 2c$. If we write that as $-9dx^3- 6cx^2+ (3d- 2b)x+ 2c= i+ jx+ kx^2+ lx^3$ then we have -9d= l, -6c= k, 3d- 2b= j, and 2c= i.

From -6c= k we have c= -k/6. Then 2c= i becomes 2(-k/6)= -k/3= i so k= -3i. That is the one condition on i, j, k, and l we have: we can write any vector in the range as $i+ jx+ (-3i)x^2+ lx^3= (1- 3x^2)i+ jx+ lx^3$. $\{1- 3x^2, x, x^3\}$ is a basis.

 

[retracell]

Sorry to bring this up again but I'm working on a similar probably and I do not understand how you found the range. At the point when you have a system of four equations 9d= l, -6c= k, 3d- 2b= j, and 2c= i, why did you choose the second and fourth equation to work with? Wouldn't the first and third also work?

Such as d=-l/9 then 3d-2b=j becomes 3(-l/9)-2b=j? What I'm trying to understand is how did you end up with k=-3i being the only condition.

 

[HallsofIvy]

Sorry to bring this up again but I'm working on a similar probably and I do not understand how you found the range. At the point when you have a system of four equations 9d= l, -6c= k, 3d- 2b= j, and 2c= i, why did you choose the second and fourth equation to work with? Wouldn't the first and third also work?

Such as d=-l/9 then 3d-2b=j becomes 3(-l/9)-2b=j? What I'm trying to understand is how did you end up with k=-3i being the only condition.

3(-l/9)- 2b= 2j does not help you because you still have "b". You want to find necessary relations between i, j, k, and l only. I used the second and fourth equations because they both involve only "c" and so "c" can be eliminated: -6c= k and 2c= i so, mulitplying the last equation by -3, -6c= -3i= k. That is the only pair of equations in which all of a, b, c, and d can be eliminated. I also know that I can only find one condition because I had alread found that the nullity was 1 so the rank must be 4- 1= 3. There can only be one condition to reduce the dimension of 4 to 3.