Sherman-Morrison-Woodbury formula
Luan V Nguyen
Can anyone of you help me for this statement:
U and V are n-by-k matrices, A is an n-by-n and is invertible.
Prove that:
A + UV^t is invertible iff I + V^t A^{-1}U is invertible.
Note: V^t is V transpose
A^{-1} is A inverse
This statement is the first part of the Sherman-Morrison-Woodbury
formula. Proving the formula itself is easy, but for this part, I've
given up :-(
Any help would be appreciated very much
Michael
Charles Crawford
x + A^(-1)UV^tx = 0
V^tx + V^tA^(-1)UV^tx = 0
Let y = V^tx
y + V^tA^(-1)Uy = 0
V^tx cannot be zero since Ax is not zero.
-- Chuck Crawford, Toronto
klimpelt
> This statement is the first part of the Sherman-Morrison-Woodbury
Can anyone give me a histoty of the Sherman-Morrison-Woodbury formula.
I once read something in a book about sparse matrices, that it was
found in about
1950, later been generalised, and that 1973, a conection to sparse
matrices was found
The Formula gives the inverse of something like
M=A+USV
The conection found in 1973 goes something like this
Ax+USVx=b
is equivalent to
Ax+Uy = b
Vx-S^-1y = 0
when you write this as a matrix (H=-S^-1) equation,
(A U) (x) = (b)
(V H) (y) = (0)
and use Gaussian elimination on the blocks, you obtain the formula.
Does Sherman,Morrison and Woodbury have really found the formula,
without being aware of this interpretation? (Or at least without
publishing it?)
Any information would be appreciated very much
Thomas
Jun Zhang
"Updating the Inverse of A Matrix",
SIAM Review, Vol. 31, No. 2, pp. 221-239, 1989.
The paper contains some historic accounts, including
the typographic errors of the original paper.
Jun Zhang
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In article <38148E87...@hivaoa.unice.fr>,
--
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* Jun Zhang * E-mail: jzh...@cs.uky.edu *
* Department of Computer Science * URL:http://www.cs.uky.edu/~jzhang *
* University of Kentucky * Tel:(606)257-3892 *