# Finding kernel and range for polynomials transformation

• charmmy
I could have used, as you say, the first and third equations: 9d= l and 3d- 2b= j. But then I would have to eliminate "d" instead of "c" which would have been more complicated. I would have to eliminate "d" from 9d= l and 3d- 2b= j to get a condition involving only a, b, and c. Doing it the other way, I can eliminate "c" immediately. And, as it turned out, I could eliminate "j" and "k" too.
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??

## The Attempt at a Solution

Last edited:
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.

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.

retracell said:
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.

## What is a polynomial transformation?

A polynomial transformation is a mathematical process that involves taking a polynomial function and applying a set of rules or operations to it, resulting in a new polynomial function. This new function may have a different shape, size, or position compared to the original function.

## How do I find the kernel of a polynomial transformation?

The kernel of a polynomial transformation is the set of all inputs (or independent variables) that result in an output (or dependent variable) of zero. To find the kernel, set the polynomial function equal to zero and solve for the independent variable. The solutions to this equation will be the elements of the kernel.

## What is the range of a polynomial transformation?

The range of a polynomial transformation is the set of all possible outputs (or dependent variables) that can be obtained by plugging in different inputs (or independent variables) into the polynomial function. To find the range, you can either graph the polynomial function and identify all possible y-values, or use algebraic techniques to determine the range.

## How can I determine the degree of a polynomial transformation?

The degree of a polynomial transformation is the highest exponent or power of the polynomial function. To determine the degree, look at the terms with the highest exponents in the polynomial function. The sum of the exponents in these terms will be the degree of the polynomial.

## What is the difference between a linear and quadratic polynomial transformation?

A linear polynomial transformation is a polynomial function with a degree of 1. This means that the highest exponent in the function is 1, resulting in a straight line when graphed. On the other hand, a quadratic polynomial transformation is a polynomial function with a degree of 2, resulting in a curved shape when graphed. It has a higher degree and therefore can produce more complex curves compared to a linear polynomial transformation.

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