Question

Let, A=(3 4)^t

1. Find the Moore-Penrose-inverse of A and A^T
2. Find the pseudo normal solutions of Ax=[2 0]^T and A^T y= 2
3. Explain results geometrically

Answer 1

very tough

Answer 2

1. The MP inverse of A^+ = [3/25 4/25], a row vector. The MP inverse of its transpose is the same vector, but it is a column.
2. The "pseudo-normal" solution to Ax = [2 0]^T is x = A^+[2 0]^T = 6/25 The particular solution to A^T y = 2 is y = [6/25 8/25]^T, a column vector.
3. The first solution is the least squares solution to Ax = b; the second solution is the vector of minimum norm in the solution space.

Answer 3

Thank you mustantiger...Could you explain the algorithms that was used for the solution?

Answer 4

A^+ = (A^TA)^(-1) A^T, when that inverse exits. Then I know the properties of the pseudo inverse. The pseudo inverse of a non-zero vector is easy: it is the corresponding row vector divided by the sum of squares.