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Question: In the accompanying figure, we show the kernel and the image of a linear transformation Lfrom R2 to R2 , together with some vectors. v→1,w→1,w→2,w→3We are told that L(v→1)=w→1. For i = 1,2,3 find the vectors L+(w→i), where L+s the pseudoinverse of Ldefined in Exercise 13. Show your solutions in the figure, and explain how you found them.

Short Answer

Expert verified

L+(w→2)=L+(projim(L)w→2)=L+(0→)=0→L+(w→3)=L+(projim(L)w→3)

Step by step solution

01

Minimal solution of Ker (L).

Since L+(w→1)is the minimal solution of the system . Thus, from Exercise 9, the minimal solution is perpendicular to Ker (L).

Consider the solution below

L+(w→2)=L+(projim(L)w→2)=L+(0→)=0→L+(w→3)=L+(projim(L)w→3)

Consider the diagram below.L+x→=w→

Hence,

L+(w→2)=L+(projim(L)w→2)=L+(0→)=0→L+(w→3)=L+(projim(L)w→3)inthediagram.

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