Linear Algebra Null Space and Range

[Punkyc7]

give a basis for the range and the null space of T:P2(R) to P1(R)
where for all p element of P2(R), T(p)=3p'' - p'


I got the null space is {1} and the range is {x,x^2} but the answer says it should be {1,x} for the range. How can something be apart of the null space and the range if its not the zero vector?

 

[Mark44]

The range can't possibly include x² since you are given that T maps polynomials of degree <= 2 to polynomials of degree <= 1.

If p(x) = ax² + bx + c, then T(p) = -2ax + 6a - b. The only way T(p) = 0 for all x is if both a and b are zero.

 

[HallsofIvy]

The null space and range are in different spaces so have nothing to do with one another (except that their dimensions add to the dimension of the domain space). Yes, if y is a constant y''- y'= 0 so the null space is spanned by {1} (I would not say the null space is {1}- that's just a single vector). If y is in the range, then p''- p'= y. If p= ax^2+ bx+ c then 3p&#039;&#039;-p&#039;= 6a+ 2ax+ b= 2ax+ (6a+b)

 

[Mark44]

The null space and range are in different spaces so have nothing to do with one another (except that their dimensions add to the dimension of the domain space). Yes, if y is a constant y''- y'= 0 so the null space is spanned by {1} (I would not say the null space is {1}- that's just a single vector). If y is in the range, then p''- p'= y. If p= ax^2+ bx+ c then p&#039;&#039;-p&#039;= 2a+ 2ax+ b= 2ax+ (2a+b)

It's given that T(p) = 3p'' - p', not p'' - p'. This affects the calculation of the range.