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## Homework Statement

Let T: P

_{4}--->P

_{3}be a linear transformation given by T(p)=p'. What is the kernel of T?

## Homework Equations

## The Attempt at a Solution

T(a

_{0}+a

_{1}+a

_{2}x

^{2}+a

_{3}x

^{3}+a

_{4}x

^{4})=a

_{1}+2a

_{2}x+3a

_{3}x

^{2}+4a

_{4}x

^{3}

Ker(T)= { T(p)=0}

so, a

_{1}+2a

_{2}x+3a

_{3}x

^{2}+4a

_{4}x

^{3}=0

then a

_{1}=2a

_{2}x+3a

_{3}x

^{2}+4a

_{4}x

^{3}

Ker(T)= { (-2,1,0,0), (-3,0,1,0), (-4,0,0,1)}

I solved this based off an example from class. But when I checked the dim[Ker(T)]= 3 and the dim[Rng(T)]=4 since my Rng(T)= {1,x,x

^{2},x

^{3}} and dim[P

_{4}]=5

using general rank nullity theorem I have 7=5, which doesn't make sense. So i'm wondering where I went wrong.

Thank you for your help.