by using R A7×4 = U?V T with singular values ?1 =12,?2 =9,?3 =5, and ?4 =2. The

Question

by using R

A7×4 = U?V T with singular values ?1 =12,?2 =9,?3 =5, and ?4 =2. The matrix U has column vectors u1,u2,u3,u4,u5,u6,u7 and the matrix V has column vectors v1,v2,v3.v4. 1. (8) Ax = y, where y =7u1 +4u4. Express x as a linear combination of the column vectors of V. 2. (8) x =2v1?v3+5v4. Express the vector Ax as a linear combination of the column vectors of U. 3. (12) Ax = b, where b =4u4. What is the norm of the error for the least squares solution? 4. (24) Ax = b, where b =3u1 ?5u2 +7u3 +u4 ?3u5 +12u6 +4u7. What is norm of the error for the least squares solution? Express the least squares solution as a linear combination of the column vectors of V. 5. (20) When solving the least squares problem Am×nx = b, we can express any vector b as

. Suppose b and U are given. How do you find the values for the ci constants? A7×4 = U?V T with singular values ?1 = 7,?2 = 4,?3 = 3, and ?4 = 0. The matrix U has column vectors u1, u2, u3, u4, u5, u6, u7 and the matrix V has column vectors v1, v2, v3.v4. Note that A is rank deficient. 6.(12) Ax = b, where b = u3 + 4u4 + 3u6. What is the norm of the error for the least squares solution? 7. (16) For the previous problem, express the solution with minimum norm as a linear combination of the column vectors of V . How many solutions are there for this least squares problem?

Explanation / Answer

1. Ax = y, where y = 7u1 +4u4.

=> x = 2v1 v3 + 5v4

2. Given x = 2v1 v3 + 5v4

We know that Ax = y

=> Ax = y = 7u1 +4u4.

3. Ax = b, where b = 4u4

We know that Ax = 7u1 + 4u4

=> Ax = 7u1 + b

So the norm of the error for the least squares solution is 7u1.

4. Ax = b, where b = 3u1 5u2 + 7u3 + u4 3u5 + 12u6 + 4u7

We know that Ax = 7u1 + 4u4

The norm of the error for the least squares solution = |Ax - b| = |7u1 + 4u4 - 3u1 5u2 + 7u3 + u4 3u5 + 12u6 + 4u7|

=> |Ax - b| = 4u1 - 5u2 + 7u3 + 5u4 - 3u5 + 12u6 + 4u7

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