Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Woodbury matrix identity
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== Special cases === When <math>V, U</math> are vectors, the identity reduces to the [[Sherman–Morrison formula]]. In the scalar case, the reduced version is simply <math display="block">\frac{1}{1 + uv} = 1 - \frac{uv}{1 + vu}.</math> ==== Inverse of a sum ==== If ''n'' = ''k'' and ''U'' = ''V'' = ''I''<sub>''n''</sub> is the identity matrix, then <math display="block">\begin{align} \left(A + B\right)^{-1} &= A^{-1} - A^{-1} \left(B^{-1} + A^{-1}\right)^{-1} A^{-1} \\[1ex] &= A^{-1} - A^{-1} \left(A B^{-1} + {I}\right)^{-1}. \end{align}</math> Continuing with the merging of the terms of the far right-hand side of the above equation results in [[Hua's identity]] <math display="block">\left({A} + {B}\right)^{-1} = {A}^{-1} - \left({A} + {A}{B}^{-1}{A}\right)^{-1}.</math> Another useful form of the same identity is <math display="block">\left({A} - {B}\right)^{-1} = {A}^{-1} + {A}^{-1}{B}\left({A} - {B}\right)^{-1},</math> which, unlike those above, is valid even if <math>B</math> is [[singular matrix|singular]], and has a recursive structure that yields <math display="block">\left({A} - {B}\right)^{-1} = \sum_{k=0}^{\infty} \left({A}^{-1}{B}\right)^k{A}^{-1}</math> if the [[spectral radius]] of <math>A^{-1}B</math> is less than one. That is, if the above sum converges then it is equal to <math>(A-B)^{-1}</math>. This form can be used in perturbative expansions where ''B'' is a perturbation of ''A''.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)