Are Linear Transformations from P2 to P3 Solvable with Polynomial Knowledge?

[mpm]

Can anyone at least tell me how to get started on this problem I have?

Problem:
Determine whether the following are linear transforatmions from P2 to P3.
L(p(x)) = xp(x)

I understand when it's in vector form but not really picking up on the polynomial part of this.

 

[calvino]

Let me start by giving you the definition of a linear transformation, in case you didnt already know.

A function L: R^n--->R^m is called a linear transformation or linear map if it satisfies

i) T(u+v)= T(u) + T(v) for all u,v in R^n
ii) T(cv)= cT(v) for all v in R^n, and scalar c

How does that help?

 

[calvino]

Sorry...I seemed to have used notation wrongly. I'm use to calling T my transformation, so I wrote my definition in such a way. I really mean..

i) L(u+v)= L(u) + L(v) for all u,v in R^n
ii) L(cv)= cL(v) for all v in R^n, and scalar c

 

[mpm]

Yes I knew the two statements that you wrote. I guess I may just be a little unclear on how to handle this problem as far as notation goes.

I have some work and an answer. Does this look sound?

L(a*p(x) + b*p(x)) = L(a*x*p(x)) + L(b*x*p(x)) = a*x*L(p(x)) + b*x*L(p(x))

Therefore this is a linear transformation.



I know this is a linear transform but I just want to make sure my proof is right.

 

[calvino]

Well, let's try this one step at a time.

i) If we take two polynomials p'(x) and p''(x) in P2, and multiply them each by x...what do we get? I think this is the main thing that your missing in your proof.

 

[calvino]

i meant to type "you're"*

 

[calvino]

actually...u know what?..let me start over. maybe I am getting carried away. =\

 

[mpm]

So the x is used as a constant basically which is practically the original definition?

 

[calvino]

I believe your method for proving it is on the right track. There are just a phew things wrong with it.

Firstly, in "L(a*p(x) + b*p(x))", both polynomials should noted as the same. You should probably distinguish the two, as the def'n states that states for any u,v.

secondly, we don't know this "L(a*p(x) + b*p(x)) = L(a*x*p(x)) + L(b*x*p(x))". What we know is that L(p(x)= x(p(x). So L(a*p(x) + b*p(x)) = x(a*p(x) + b*p'(x)).

perhaps you can take it from there. Sorry about the frustration earlier. I really did lose track.

 

[calvino]

also...phew=few. I'm not sure what's wrong with me.

 

[mpm]

Im curious if i can do it this way. It just came to me and I may be breaking a rule.

Can I put:

L(xp(x)) = xL(p(x)) = xp(x)

And that satisfies my question?

 

[calvino]

Im curious if i can do it this way. It just came to me and I may be breaking a rule.
Can I put:
L(xp(x)) = xL(p(x)) = xp(x)
And that satisfies my question?

xL(p(x))=x(x(px)), by def'n of L(p(x)).

 

[mpm]

Ok I see what you mean. Maybe I will get this straight sometime. I am terrible at linear algebra. Calculus and differential equations are more my thing.

 

[mpm]

Ok I think I might have it this time. Can someone tell me if I am correct?


L(a(p(x)) = x(ap(x)) = a(xp(x)) = aL(p(x))

Does this look correct?

 

[mpm]

I also forgot to add this part to prove the second part.

L(p(x) + p'(x)) = x(p(x) + p'(x)) = xp(x) + xp'(x) = L(p(x)) + L(p'(x))

I think this should be right but would like to be corrected if wrong.

 

[TD]

Both are right as far as I can see.
Note that you can always replace the two statements that you check by just one, which is usually faster. For more complex transformation, it may be better to prove the two statements apart, but for easier ones: just show that "L of a lineair combination gives a lineair combination of L's"

$$L\left( {ap\left( x \right) + bq\left( x \right)} \right) = x\left( {ap\left( x \right) + bq\left( x \right)} \right) = axp\left( x \right) + bxq\left( x \right) = aL\left( {p\left( x \right)} \right) + bL\left( {q\left( x \right)} \right)$$