# Kernal and Range of a Linear Transformation

1. ### hkus10

50
Let L:p2 >>> p3 be the linear transformation defined by L(p(t)) = t^2 p'(t).
(a) Find a basis for and the dimension of ker(L).
(b) Find a basis for and the dimension of range(L).

The hint that I get is to begin by finding an explicit formula for L by determining
L(at^2 + bt + c).
Does this hint mean let p(t) = at^2 + bt + c?
Then, I find that t^2 p'(t) = 2at^3 + bt^2.
Then, I conclude that the basis for ker(L) = {1}.
Is it right?
Also, how to find range(L)?

Thanks

2. ### hunt_mat

1,613
Factorise...

$$L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b)$$

This will be zero when t=0 or t=-b/2a, so...

3. ### Dick

25,887
That doesn't have much to do with the problem. hkus10 correctly has ker(L)={1} and having written L(p(t))=2at^3+bt^2 the answer to the dimension and a basis of range(L) should be pretty obvious. Why isn't it hkus10? What's a basis for p3?

4. ### hkus10

50
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?

5. ### Dick

25,887
Two steps forward, one step backward. Yes, that's a basis for range(L). But now your basis for ker(L) is wrong. I liked your ker(L)={1} a lot better. Why did you put t in? Is t in the kernel?

6. ### Char. Limit

1,986
I believe you have a workable basis for range(L). However, I think your basis for ker(L) has too many entries.

7. ### hkus10

50
Since L(at^2 + bt + c) = 2at^3 + bt^2
No matter what value of t and 1, 2a^3 + bt^2 should always give me 0 vector. So, I have a question why Ker(L) does not have t as a basis?
Another question is dim Ker(L) + dim range(L) = dim (p3) by thm. since the dim range(L) = 2 and dim (p3) = 4, why dim ker(L) not equal to 2?

8. ### Dick

25,887
The rank nullity theorem tells you dim ker(L)+dim range(L)=dim(p2). Not dim(p3). dim(p2)=3. t is not in ker(L) because L(t) is not zero. L(t)=t^2. t^2 is not zero.

50
Thanks