Register to reply

Kernal and Range of a Linear Transformation

by hkus10
Tags: kernal, linear, range, transformation
Share this thread:
hkus10
#1
Apr21-11, 04:38 PM
P: 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
Phys.Org News Partner Science news on Phys.org
Scientists discover RNA modifications in some unexpected places
Scientists discover tropical tree microbiome in Panama
'Squid skin' metamaterials project yields vivid color display
hunt_mat
#2
Apr21-11, 05:58 PM
HW Helper
P: 1,583
Factorise...

[tex]
L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b)
[/tex]

This will be zero when t=0 or t=-b/2a, so...
Dick
#3
Apr21-11, 06:20 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
Quote Quote by hunt_mat View Post
Factorise...

[tex]
L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b)
[/tex]

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

hkus10
#4
Apr21-11, 09:59 PM
P: 50
Kernal and Range of a Linear Transformation

is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?
Dick
#5
Apr21-11, 10:08 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
Quote Quote by hkus10 View Post
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?
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?
Char. Limit
#6
Apr21-11, 10:12 PM
PF Gold
Char. Limit's Avatar
P: 1,956
Quote Quote by hkus10 View Post
is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?
I believe you have a workable basis for range(L). However, I think your basis for ker(L) has too many entries.
hkus10
#7
Apr22-11, 06:58 AM
P: 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?
Dick
#8
Apr22-11, 07:12 AM
Sci Advisor
HW Helper
Thanks
P: 25,235
Quote Quote by hkus10 View Post
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?
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.
hkus10
#9
Apr22-11, 08:32 AM
P: 50
Quote Quote by Dick View Post
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.
Thanks


Register to reply

Related Discussions
Range of a Matrix Transformation linear algebra Calculus & Beyond Homework 5
Having difficulty understanding what the Range of a linear transformation is. Linear & Abstract Algebra 3
Linear algebra - Kernal and Range Calculus & Beyond Homework 7
Kernal, range and linear transformations Calculus & Beyond Homework 5
Kernal and Range (share zero vector?) Calculus & Beyond Homework 13