Question

Suppose A = UΣV ⊤ ∈ R m×n is a SVD decomposition. Consider AZ = B for B ∈ R m×p . For what conditions does AZ = B have a (1) solution, (2) unique solution, (3) at most one solution? If there exists a solution, use U, Σ, V to represent it.

Answer 1

The conditions for the equation AZ = B to have a solution, unique solution, or at most one solution are determined by the properties of the singular value decomposition (SVD) of matrix A. If B lies in the column space of A, the equation has a solution. If A is full rank, the equation has a unique solution. If A is not full rank, the equation may have at most one solution. The solution can be represented using the SVD decomposition of A.

Step-by-step explanation:

To determine the conditions under which the equation AZ = B has a solution, unique solution, or at most one solution, we must consider the properties of the singular value decomposition (SVD). The SVD decomposition of matrix A is given by A = U Σ V^T, where U and V are orthogonal matrices and Σ is a diagonal matrix containing the singular values of A.

1. If and only if the matrix B lies in the column space of A, AZ = B will have a solution. This means that B can be expressed as a linear combination of the columns of A.
2. The equation AZ = B will have a unique solution if and only if the matrix A is full rank, meaning that all the singular values in Σ are non-zero.
3. If A is not full rank, the equation AZ = B may have at most one solution. This occurs when B lies in the column space of A, but there are multiple vectors in the null space of A that satisfy the equation.

To represent the solution, we can use the SVD decomposition of A. We can express B as B = U Σ V^T Z, where Z is a matrix of appropriate dimensions. Then, the solution C can be obtained by multiplying B by the matrix V Σ^(-1) U^T.

Learn more about Solving linear equations