Question: 9. Suppose we are solving for Ax = b with AERmXn being rank deficient (minimum norm solution). [12] (a) For b&R(A) why wouldn't the solution

9. Suppose we are solving for Ax = b with AERmXn being rank deficient (minimum norm solution). [12] (a) For b&R(A) why wouldn't the solution (ATA)-1 ATb work? (b) Suppose bER(A). Show that the error vector is indeed zero in this case. (c) Say, BER(A). Verify that the minimum norm solution that we obtained in the class is indeed a solution. Also show that the choice of null space basis vectors from N (A) and N (AT) doesn't affect the error. (d) Comment on the uniqueness of SVD in light of your answer to the above question. (e) Let r be the rank of A and let r - 1 singular values repeat. What can you say about uniqueness of SVD in this case? 9. Suppose we are solving for Ax = b with AERmXn being rank deficient (minimum norm solution). [12] (a) For b&R(A) why wouldn't the solution (ATA)-1 ATb work? (b) Suppose bER(A). Show that the error vector is indeed zero in this case. (c) Say, BER(A). Verify that the minimum norm solution that we obtained in the class is indeed a solution. Also show that the choice of null space basis vectors from N (A) and N (AT) doesn't affect the error. (d) Comment on the uniqueness of SVD in light of your answer to the above question. (e) Let r be the rank of A and let r - 1 singular values repeat. What can you say about uniqueness of SVD in this case
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