Verify Linear Transformations: 3.4a, 3.4b

[laminatedevildoll]

I've uploaded a document which I am currently working on. I would like to verify if I am doing these problems correctly. Thank you.

In the first attachment (3.4b)
For 1.
a. 4x^3-2x
b. T(P)=0
ker T={C:C \inR}
Im T = {P|P is less than degree 3 or less}

c. T is not one to one because P is a constant. T is not onto because it's degree less than 3. I am not sure if I am proving this right. I'd appreciate some help.

For 2.
a. 1/5x^5-1/3x^3+C
b.
ker T={C:C \inR}
Im T = {P|P is less than degree 5 or less}
T is not one to one because C=0. But, T is on-to. How do I prove this?

In the second attachment (3.4a) part 4.
I proved that T is one to one because T(f)=T(g), f=g How do I prove that this is on-to?

 

[laminatedevildoll]

nvm, I understand it.

 

[M98Ranger]

Overall, your work looks good and your reasoning is correct. Here are some suggestions for further clarifications:

For 3.4b, 1:
a. Your answer for T(P) is correct, but you could also write it as T(P) = 0x^3 + 0x^2 + 0x - 0. This might make it clearer that T(P) is a polynomial of degree less than or equal to 3.
b. Your answer for ker T and Im T are correct, but you could also add a brief explanation for why they are the same.
c. Your reasoning is correct, but you could also mention that a constant polynomial cannot be mapped to a non-constant polynomial, so T is not onto.

For 3.4b, 2:
a. Your answer for T(P) is correct, but you could also write it as T(P) = 1/5x^5 + (-1/3)x^3 + C. This might make it clearer that T(P) is a polynomial of degree less than or equal to 5.
b. Again, your answers for ker T and Im T are correct, but you could add a brief explanation for why they are the same.
c. To prove that T is onto, you could show that for any polynomial of degree less than or equal to 5, there exists a polynomial P such that T(P) = that polynomial. In other words, show that for any polynomial Q(x) of degree less than or equal to 5, you can find a polynomial P such that T(P) = Q(x). This would show that every polynomial of degree less than or equal to 5 is in the image of T, and thus T is onto.

For 3.4a, 4:
Your proof for T being one-to-one is correct. To show that it is onto, you could use a similar approach as in 3.4b, 2c. Show that for any polynomial Q(x), there exists a polynomial P such that T(P) = Q(x). This would show that every polynomial is in the image of T, and thus T is onto.

Overall, your work is correct and well-reasoned. Keep up the good work!