Ok.
If I then solve the system Ax =b where the matrix A (m x n) with rank m by means of the right-hand pseudo inverse A^T(A*A^T)^-1, what I get is the minimum norm solution to the m independent columns of A? Or...?
What happens to the remaining free variables? Are they in the null-space?
Sorry. From the original problem where m < n, there is not a set of linearly independent vectors to yield a exact solution to the system Ax = b. Then a common technique to solve for a vector that comes as close as possible, would be to form the pseudoinverse, right? This pseudoinverse, is that a...
Hello,
I was wondering if the pseudoinverse can be considered a change of basis?
If an m x n matrix with m < n and rank m and you wish to solve the system Ax = b, the solution would hold an infinite number of solutions; hence you form the pseudoinverse by A^T(A*A^T)^-1 and solve for x to...