Tilted Forum Project Discussion Community - View Single Post

Yakk · 05-09-2004, 02:02 AM

You might want to do what Knife did in a few more steps.

Let A be a matrix, and let D be the orthogonal diagonalization of A.
Then there exists a P such that
1/P * A * P = D
Let 1/A be the inverse of A.

Examine C = 1/P * 1/A * P.
C * D = 1/P * 1/A * P * 1/P * A * P
= 1/P * 1/A * (P * 1/P) * A * P
= 1/P * (1/A * A) * P
= 1/P * P
= I
Thus, C = 1/D
As D is diagonal, its inverse is diagonal. Thus, C is diagonal. Then, as
C = 1/P * 1/A * P
by the definition of orthogonal diagonalization, 1/A can be orthogonally diagonalized.

05-09-2004, 02:02 AM	#6 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	You might want to do what Knife did in a few more steps. Let A be a matrix, and let D be the orthogonal diagonalization of A. Then there exists a P such that 1/P * A * P = D Let 1/A be the inverse of A. Examine C = 1/P * 1/A * P. C * D = 1/P * 1/A * P * 1/P * A * P = 1/P * 1/A * (P * 1/P) * A * P = 1/P * (1/A * A) * P = 1/P * P = I Thus, C = 1/D As D is diagonal, its inverse is diagonal. Thus, C is diagonal. Then, as C = 1/P * 1/A * P by the definition of orthogonal diagonalization, 1/A can be orthogonally diagonalized. __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.