Construction of minimum norm solution matrix

[kalleC]

Homework Statement

Consider the linear system of equations Ax = b b is in the range of A
Given the SVD of a random matrix A; construct a full rank matrix B for which the solution:
x = B^-1*b
is the minimum norm solution.

Also A is rank deficient by a known value and diagonalizable

Homework Equations

The Attempt at a Solution

I am completely clueless here. I know that due to rank deficiency some eigenvalues of A are zero and this has something to do with the corresponding left and righ singular vectors but I am clueless as to what. I have read through all our notes and can find nothing relating to this problem. Any hint to get me started would be appreciated.

 

[mighty2000]

Dear fellow scientist,

I understand your confusion and I am happy to provide some guidance to help you solve this problem.

Firstly, let's review the concept of rank deficiency. A matrix A is considered rank deficient if its rank is less than its number of columns or rows. This means that A is not invertible and there are infinitely many solutions to the system Ax = b. However, we can still find a minimum norm solution to this system by using the SVD (Singular Value Decomposition) of A.

The SVD of a matrix A is given by A = UΣV^T, where U and V are orthogonal matrices and Σ is a diagonal matrix with the singular values of A on its diagonal. Since A is rank deficient, it means that there are at least one or more zero singular values on Σ.

Now, to construct a full rank matrix B for which the solution x = B^-1*b is the minimum norm solution, we need to use the pseudo-inverse of A. The pseudo-inverse of A is given by A^+ = VΣ^+U^T, where Σ^+ is the pseudo-inverse of Σ. The pseudo-inverse of Σ can be obtained by taking the reciprocal of all non-zero singular values on Σ and then taking the transpose. This means that if there are any zero singular values on Σ, they will remain as zeros on Σ^+.

Therefore, to construct B, we can simply replace the zero singular values on Σ with non-zero values. This will result in a full rank matrix B, and when we solve for x using the equation x = B^-1*b, we will get the minimum norm solution.

I hope this helps to get you started on solving this problem. Good luck!