Question

Find the kernel and range

Answer 1

I need to find the kernel and range og each of the following linear operators on P_3:

(a): \[L(p(x))=xp'(x)\]

(b)
\[L(p(x))=p(x)-p'(x)\]

(c)
\[L(p(x))=p(0)x+p(1)\]

Answer 2

What's p(x)?
Some function? Or maybe a polynomial?

Answer 3

Not that it matters, really, apparently.
Now, kernel, it's the set of all elements in your domain (and in your case, it seems that the domain consists of functions) that are mapped to zero.

Answer 4

I just have the following information

Answer 5

Granted :)
Let's do (a)
L(p(x)) = xp'(x)
So equate this to zero...
xp'(x) = 0
What do you get?

Answer 6

Stay with me here... @RFJ-86 XD

Answer 7

:)

Answer 8

xp'(x) = 0
When the product of two elements is zero, then

(assuming the field has characteristic zero, but never mind that >.< )

at least one of them has to be zero.
So either
x = 0
or
p'(x) = 0

Answer 9

And while you're thinking about that, I think \(P_3\) is the set of polynomials of degree 3.
What do you think @experimentX ?

Answer 10

p(x) = constant <-- kernet
image is a a polynomial of degree 'n' with no constant term ... i think

Answer 11

no, I meant what is \(P_3\).
We'd have a right better change at figuring out the ranges if we knew what it is :D

Answer 12

chance*
sorry

Answer 13

oh ... you meant what is P_3 , i suppose it is what you think.

Answer 14

Right. Proceeding given that then :)

Answer 15

@RFJ-86
Once again...
either
x = 0
or
p'(x) = 0

and if p'(x) = 0
what must be p(x)?

Answer 16

A constant?

Answer 17

Correct :3
So, the kernel of that operator is the set of all constants :)
(Actually, it's the set of all constants union the set {0} but then again, 0 is also a constant)

Answer 18

Can you do the next one?

Answer 19

p(x)=p'(x)

Answer 20

So p(x)=0

Answer 21

that's right :)
(assuming P_3 are polynomials of course, otherwise, an obvious candidate would be multiples of e^x ^_^ )

Answer 22

The only 'polynomial' whose derivative is itself is the zero-polynomial :)
Do the last one :)

Answer 23

p(0)x=p(1)

Answer 24

so... what's the kernel?

Answer 25

Sorry
-p(0)x=p(1)

Answer 26

Still not the kernel... that's just an equation.
Kernel is a set :P

Answer 27

Please remember that we are dealing with polynomials themselves and not with finding the value of x.

Answer 28

The kernel would be a set of certain polynomials...

Answer 29

Okay..
So
(a) --> all constant
(b) --> 0
(c) --> a set of polynomials

Answer 30

c is not yet answered.
You need to be specific as to what set of polynomials it is :P

Answer 31

hmm...

Answer 32

I cant spot the solution. :(

Answer 33

Well for p(0)x + p(1) to be the zero polynomial
both
p(0) and p(1) have to be zero.

So... it's the set of (max degree 3) polynomials with both 0 and 1 as roots.

Or the set
x(x - 1)(ax + b) for all real numbers a and b

Answer 34

Of course.. And the range?

Answer 35

range ...
the set of all possible images.

Answer 36

So for c it is all real numbers?

Answer 37

no... if you must have notation, it's the set
\[\Large \left\{x(x-1)(ax+b) \left|a,b \in \mathbb{P}\right.\right\}\]

Answer 38

Så for (a) i can write the range this way? (with {})
\[k |\in R\]
@terenzreignz

Answer 39

*so

Answer 40

the range for the first is the entire P3

Answer 41

k | k in P3

Answer 42

yeah :)
again, to stress, your ranges are sets of polynomials, not numbers.

Answer 43

Sorry but my math notation is bad.

Answer 44

:)

Answer 45

And d is {0 | 0 in P3}

Answer 46

sorry b

Answer 47

no, that's the kernel.
you need the range

Answer 48

How can I write b?

Answer 49

I'm still scratching my head as to the range :D

Answer 50

1) i think range is in p_n(x) - p_n(0), p_n(x), p_1(x) not sure how to write that.

Answer 51

@experimentX So you don't think the ranges in (c) are {x(x-1)*(ax+b)| a,b in P} or what?

Answer 52

@terenzreignz

Answer 53

the range is of the form ax + b ... a two dimensional vector space.