Finding basis for nullspace of transformation

[Clandry]

T: P2 → R (the 2 is supposed to be a subscript) The P is supposed to be some weird looking P denoting that it is a polynomial of degree 2.
T (p(x)) = p(0)

Find a basis for nullspace of linear transformation T.The answer is {x, x^2}

I want to make sure I'm interpreting this correctly.

It only goes to x^2 because of P2 right? Like if it was P3 it would be x, x^2, x^3?

Also I don't understand why it's {x, x^2} is it because x can be 0?

 

[Mark44]

T: P2 → R (the 2 is supposed to be a subscript) The P is supposed to be some weird looking P denoting that it is a polynomial of degree 2.
T (p(x)) = p(0)

Find a basis for nullspace of linear transformation T.


The answer is {x, x^2}

I want to make sure I'm interpreting this correctly.

It only goes to x^2 because of P2 right? Like if it was P3 it would be x, x^2, x^3?

Also I don't understand why it's {x, x^2} is it because x can be 0?

If p(x) = a₂x² + a₁x + a₀, what is T(p(x))?

 

[Clandry]

If p(x) = a₂x² + a₁x + a₀, what is T(p(x))?

oooo, just a0. Which must be 0 if the p(x) you give is to satisfy the requimrenets for N(T) right?

So essentially the problem is saying P(x)=x and P(x)=x^2?

 

[Mark44]

oooo, just a0. Which must be 0 if the p(x) you give is to satisfy the requimrenets for N(T) right?

No. a₀ can be any real number.

So essentially the problem is saying P(x)=x and P(x)=x^2?

No. How can P(x) be both x and x²?

P(x) is an arbitrary 2nd degree polynomial.

What does it mean for a function to be in the nullspace in this problem?

What is a basis for P₂? What does the transformation T do to each of these basis elements?

 

[Clandry]

It means it can be spanned by the vectors in the basis which is x and x^2. I think a basis for P2 is {1, x, x^2)?

In my textbookthey define N(T)]{ all v in vector space V and T(v)=0.}
So for this problem
N(T)={P(x) such that P(0)=0}
1 would be dropped out because P(0)=0 wouldn't be satisfied.

 

[Mark44]

It means it can be spanned by the vectors in the basis which is x and x^2.

Yes, but the only reason you know that the x and x² are in a basis for the nullspace of T is because you are given the answer.

I think a basis for P2 is {1, x, x^2)?

Yes.

In my textbookthey define N(T)]{ all v in vector space V and T(v)=0.}
So for this problem
N(T)={P(x) such that P(0)=0}

No. N(T) = {P(x) $\in$ P₂ such that T(P(x)) = P(0)}

1 would be dropped out because P(0)=0 wouldn't be satisfied.

What does the transformation T do to each of the three functions in your basis?

 

[Clandry]

That's where I get a bit confused.

T(1)=1
P(x)=x then p(0)=0
So
T(x)=0
T(x^2)=0
?

 

[Mark44]

That's where I get a bit confused.

T(1)=1
P(x)=x then p(0)=0
So
T(x)=0
T(x^2)=0
?

What you have above is OK - where are you confused? You now have enough information to give a basis for Null(T).

 

[Clandry]

Got it, just learned how to do this today in class.
Since
T(1)=1
T(x)=0
T(x^2)=0

They form the transformation matrix:
1 0 0
0 0 0
0 0 0
this is spanned by the vectors {[0 1 0]^t, [0 0 1]^t}
Which correspond to x, and x^2