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.
p(x)=a0+a1xS(p(x))=0⇒∫01p(x)dx=0⇒∫01(a0+a1x)dx=0⇒⇒a0+a12=0⇒p(x)=t(1−2x) − one elementNullity:dim(ker(S))=1rank(S)=dim(P1)−dim(ker(S))=2−1=1p\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=1p(x)=a0+a1xS(p(x))=0⇒∫01p(x)dx=0⇒∫01(a0+a1x)dx=0⇒⇒a0+2a1=0⇒p(x)=t(1−2x)−oneelementNullity:dim(ker(S))=1rank(S)=dim(P1)−dim(ker(S))=2−1=1
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