This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1
cn with the unitary U = [z1 z2 · · · zn ] \\
Now, consider the general case in which A may not be invertible. In this case, following the same
procedure we get {zj }n=1 which possibly contain zero vectors. And, if zj = 0, then cj > 0 is
j
arbitrary. Then, we get ˜
A=U 1
c1
1
c2
...
1
cn ˜
with U containing zero columns so possibly it is not unitary. We can ﬁx this by inserting unit
˜
vectors into the zero columns of U , complementing the other column vectors to an orthonormal
bas...
View
Full
Document
This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.
 Fall '12
 BernhardBodmann
 Math, Matrices

Click to edit the document details