Answer to Question #319281 in Linear Algebra for peaceboy

Question

EXERCISE 2: Find the rank and the nullity of the linear transformation S: p_1→ℝ given by 

     S(p(x)) = ∫_0^1p(x)dx.

Answer 1

$p\left( x \right) =a_0+a_1x\\S\left( p\left( x \right) \right) =0\Rightarrow \int_0^1{p\left( x \right) dx}=0\Rightarrow \int_0^1{\left( a_0+a_1x \right) dx}=0\Rightarrow \\\Rightarrow a_0+\frac{a_1}{2}=0\Rightarrow p\left( x \right) =t\left( 1-2x \right) \,\,-\,\,one\,\,element\\Nullity: dim\left( ker\left( S \right) \right) =1\\rank\left( S \right) =dim\left( P_1 \right) -dim\left( ker\left( S \right) \right) =2-1=1$