QR Factorization: Show A=LQ, L Triangular & Q Orthogonal

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Homework Statement


Consider an invertible n x n matrix A. Can you write A as A=LQ, where L is a lower triangular matrix and Q is orthogonal? Hint: Consider the QR factorization of #A^T#.

Homework Equations


For QR factorization, Q is orthogonal and R is upper triangular.

The Attempt at a Solution


If we consider the hint, then we can write:
##A^T=S*U## where S is orthogonal matrix and U is some upper triangular matrix.
##(A^T)^T=U^T*S^T##; transpose of upper triangular matrix U is some lower triangular matrix L
##A=L*S^T##

Here is where I get lost. I don't know how to show that S^T=Q. Could someone please give me a hint?
 
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Clandry said:

Homework Statement


Consider an invertible n x n matrix A. Can you write A as A=LQ, where L is a lower triangular matrix and Q is orthogonal? Hint: Consider the QR factorization of #A^T#.

Homework Equations


For QR factorization, Q is orthogonal and R is upper triangular.

The Attempt at a Solution


If we consider the hint, then we can write:
#A^T+=S*U# where S is orthogonal matrix and U is some upper triangular matrix.
#(A^T)^T=U^T*S^T#; transpose of upper triangular matrix U is some lower triangular matrix L
#A=L*S^T#

Here is where I get lost. I don't know how to show that S^T=Q. Could someone please give me a hint?
Your # symbols are cluttering up your work, making it harder to read than it should be. If you trying to use LaTeX, use two # at the beginning and two more at the end.

Regarding your question, S is orthogonal, right. What about its transpose, ST? Isn't that orthogonal as well?
 
Mark44 said:
Your # symbols are cluttering up your work, making it harder to read than it should be. If you trying to use LaTeX, use two # at the beginning and two more at the end.

Regarding your question, S is orthogonal, right. What about its transpose, ST? Isn't that orthogonal as well?
Ah yes! I think I overthought this problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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