Let A be the nxn matrix whose elements in position (i, i+1) are equal to 1, , while all other elements are zero. Let b = . Show that the solution to Ax=b does not lie in for any .
We've shown in class that if Ax=b with nonsingular matrix A, the solution x always lies in some Krylov space. The A in this example, however, is being used to show that this is not always true when the matrix A is singular. I am so lost on where to begin. Does anyone have any experience with Krylov Subspaces? Thank you so much.