a) To graph: the ker(A),(kerA)⊥andim(AT) b) To find: the relationship between im (AT) and ker (A). c) To find: the relationship between ker(A) and solution set S d) To find x→0 at the intersection of ker(A)and(kerA)⊥ e) To find: the lengths of x→0 compared to the other vectors in S

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Corben Pittman · Accepted Answer

a) Given: The image of linear transformation T(x―)=A(x―) is the span of the column vectors of A. The kernel of a linear transformation T(x―)=A(x―)omRm→Rn consists of all zeros of the transformation. That is the solution of the equation T(x―)=A(x―)=0. We denote the kernel of T by ker (A) or ker (T) and it is the solution set of the linear system A(x―)=0. Graph:  Interpretation: Consider the linear system A(x→)=b→forA=[1326]andb→=[1020]. A(x→)=0→ [1326][x1x2]=0 x1=−3t,x2=t t=parametr [x1x2]=[−3tt]=[−31]t Thus, ker (A) is the line spanned by [−31]∈R2. Consider im (A) which is spanned by u→1=[12]. Then, (kerA)⊥=ker(A)−proj(kerA) (kerA)⊥=ker(A)−(u→1.kerA)u→1 (kerA)⊥=[−31]−[[12]⋅[−31]][12] (kerA)⊥=[−31]+[12] (kerA)⊥=[−23] Consider AT=[1236]. Since [1236]=

a) To graph: the ker(A),ker(A)^bot and im(A^T) b) To find: the relationship between im(A^T) and ker(A).

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