View source for Woodbury matrix identity

{{Use American English|date = January 2019}}
{{Short description|Theorem of matrix ranks}}
In [[mathematics]] (specifically [[linear algebra]]), the '''Woodbury matrix identity''', named after Max A. Woodbury<ref>Max A. Woodbury, ''Inverting modified matrices'', Memorandum Rept. 42, Statistical Research Group, Princeton University, Princeton, NJ, 1950, 4pp {{MR|38136}}</ref><ref>Max A. Woodbury, ''The Stability of Out-Input Matrices''. Chicago, Ill., 1949. 5 pp. {{MR|32564}}</ref> says that the inverse of a rank-''k'' correction of some [[matrix (mathematics)|matrix]] can be computed by doing a rank-''k'' correction to the inverse of the original matrix. Alternative names for this formula are the '''matrix inversion lemma''', '''Sherman–Morrison–Woodbury formula''' or just '''Woodbury formula'''. However, the identity appeared in several papers before the Woodbury report.<ref name="hager">{{cite journal
 |first=William W. |last=Hager
 |title=Updating the inverse of a matrix
 |journal=SIAM Review
 |volume=31 |year=1989 |pages=221&ndash;239 |issue=2
 |doi=10.1137/1031049 |mr=997457 | jstor = 2030425
}}</ref>

The Woodbury matrix identity is<ref name="higham">{{Cite book | last1=Higham | first1=Nicholas | author1-link=Nicholas Higham | title=Accuracy and Stability of Numerical Algorithms | publisher=[[Society for Industrial and Applied Mathematics|SIAM]] | edition=2nd | isbn=978-0-89871-521-7 | year=2002 | page=258 |mr=1927606 | postscript=<!--None--> }}
</ref>
:<math> \left(A + UCV \right)^{-1} = A^{-1} - A^{-1}U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1}, </math>

where ''A'', ''U'', ''C'' and ''V'' all denote matrices of the correct ([[conformable matrix|conformable]]) sizes.  Specifically, ''A'' is ''n''-by-''n'', ''U'' is ''n''-by-''k'', ''C'' is ''k''-by-''k'' and ''V'' is ''k''-by-''n''. This can be derived using [[Invertible matrix#Blockwise inversion|blockwise matrix inversion]].

While the identity is primarily used on matrices, it holds in a general [[Ring (mathematics)|ring]] or in an [[Ab-category]].

== Discussion ==
The equation above is over-parametrized. There is no loss of generality
by replacing ''A'' and ''C'' with the identity matrix ''I'':
:<math> \left(I + UV \right)^{-1} = I - U \left(I + VU \right)^{-1} V. </math>
To recover the original equation from this ''reduced identity'', set <math>U = A^{-1}X</math> and <math>V = CY</math>.

This identity itself can be view as the combination of two simpler identities
: <math> (I + P)^{-1} = I - (I + P)^{-1} P = I - P (I + P)^{-1}</math>
and the so-called '''push-through identity'''<ref name="HS"/>
: <math> (I + UV)^{-1} U = U (I + VU)^{-1}. </math>

=== Special cases ===

When <math>V, U</math> are vectors, the identity reduces to the [[Sherman–Morrison formula]]. 

In the scalar case it (the reduced version) is simply
: <math>\frac{1}{1 + uv} = 1 - \frac{uv}{1 + uv}.</math>

==== Inverse of a sum ====

If ''p'' = ''q'' and ''U'' = ''V'' = ''I''<sub>''p''</sub> is the identity matrix, then

:<math>
\begin{align}
\left({A} + {B}\right)^{-1} &= A^{-1} - A^{-1} (B^{-1} + A^{-1})^{-1} A^{-1}\\
&= {A}^{-1} - {A}^{-1}\left({I} + {B}{A}^{-1}\right)^{-1}{B}{A}^{-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>\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>\left({A} - {B}\right)^{-1} = {A}^{-1} + {A}^{-1}{B}\left({A} - {B}\right)^{-1},</math>

which has a recursive structure that yields
:<math>\left({A} - {B}\right)^{-1} = \sum_{k=0}^{\infty} \left({A}^{-1}{B}\right)^k{A}^{-1}.</math>

This form can be used in perturbative expansions where ''B'' is a perturbation of ''A''.

=== Variations ===

==== Binomial Inverse Theorem ====

If ''A'', ''U'', ''B'', ''V'' are matrices of sizes ''p''×''p'', ''p''×''q'', ''q''×''q'', ''q''×''p'', respectively, then
:<math>
  \left(A + UBV\right)^{-1} =
  A^{-1} - A^{-1}UB\left(B+BVA^{-1}UB\right)^{-1}BVA^{-1}
</math>

provided ''A'' and ''B'' + ''BVA''<sup>−1</sup>''UB'' are nonsingular. Nonsingularity of the latter requires that ''B''<sup>−1</sup> exist since it equals {{nowrap|''B''(''I'' + ''VA''<sup>−1</sup>''UB'')}} and the rank of the latter cannot exceed the rank of ''B''.<ref name=HS>{{cite journal | last1 = Henderson | first1 = H. V. | last2 = Searle | first2 = S. R. | year = 1981 | title = On deriving the inverse of a sum of matrices | url = http://ecommons.cornell.edu/bitstream/1813/32749/1/BU-647-M.pdf| journal = SIAM Review | volume = 23 | issue = | pages = 53–60 | doi = 10.1137/1023004 | jstor = 2029838 }}</ref>

Since ''B'' is invertible, the two ''B'' terms flanking the parenthetical quantity inverse in the right-hand side can be replaced with {{nowrap|(''B''<sup>−1</sup>)<sup>−1</sup>,}} which results in the original Woodbury identity.

A variation for when ''B'' is singular and possibly even non-square:<ref name=HS/>

:<math>(A + UBV)^{-1} = A^{-1} - A^{-1}U(I + BVA^{-1}U)^{-1}BVA^{-1}.</math>

Formulas also exist for certain cases in which ''A'' is singular.<ref>Kurt S. Riedel, "A Sherman–Morrison–Woodbury Identity for Rank Augmenting Matrices with Application to Centering", ''SIAM Journal on Matrix Analysis and Applications'', 13 (1992)659-662, {{doi|10.1137/0613040}} [http://math.nyu.edu/mfdd/riedel/ranksiam.ps preprint] {{MR|1152773}}</ref>

== Derivations ==

=== Direct proof ===
The formula can be proven by checking that <math>(A + UCV)</math> times its alleged inverse on the right side of the Woodbury identity gives the identity matrix:

: <math>\begin{align}
      & \left(A + UCV \right) \left[ A^{-1} - A^{-1}U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1} \right] \\
  ={} & \left\{ I - U\left(C^{-1} + VA^{-1}U \right)^{-1}VA^{-1} \right\} + \left\{ UCVA^{-1} - UCVA^{-1}U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1} \right\} \\
  ={} & \left\{ I + UCVA^{-1} \right\} - \left\{ U\left(C^{-1} + VA^{-1}U \right)^{-1}VA^{-1} + UCVA^{-1}U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1} \right\} \\
  ={} & I + UCVA^{-1} - \left(U + UCVA^{-1}U\right) \left(C^{-1} + VA^{-1}U\right)^{-1}VA^{-1} \\
  ={} & I + UCVA^{-1} - UC \left(C^{-1} + VA^{-1}U\right) \left(C^{-1} + VA^{-1}U\right)^{-1}VA^{-1} \\
  ={} & I + UCVA^{-1} - UCVA^{-1} \\
  ={} & I.
\end{align}</math>

=== Alternative proofs ===

{{collapse top|title=Algebraic proof }}
First consider these useful identities,
: <math>\begin{align}
               U + UCV A^{-1} U &= 
               UC \left(C^{-1} + V A^{-1} U\right) = \left(A + UCV\right) A^{-1} U \\
  \left(A + UCV\right)^{-1} U C &= A^{-1}U \left(C^{-1} + VA^{-1} U\right) ^{-1}
\end{align}
</math>

Now, 
: <math>\begin{align}
  A^{-1} &= \left(A + UCV\right)^{-1}\left(A + UCV\right) A^{-1}\\
         &= \left(A + UCV\right)^{-1}\left(I + UCVA^{-1}\right) \\
         &= \left(A + UCV\right)^{-1} + \left(A + UCV\right)^{-1} UCVA^{-1} \\ 
         &= \left(A + UCV\right)^{-1} + A^{-1} U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1}.
\end{align}</math>
{{collapse bottom}}

{{collapse top|title=Derivation via blockwise elimination}}
Deriving the Woodbury matrix identity is easily done by solving the following block matrix inversion problem
:<math>
  \begin{bmatrix} A & U \\ V & -C^{-1} \end{bmatrix}\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} I \\ 0 \end{bmatrix}.
</math>

Expanding, we can see that the above reduces to 

:<math>\begin{cases} AX + UY = I \\ VX - C^{-1}Y = 0\end{cases}</math>

which is equivalent to <math>(A + UCV)X = I</math>. Eliminating the first equation, we find that <math>X = A^{-1}(I - UY)</math>, which can be substituted into the second to find <math>VA^{-1}(I - UY) = C^{-1}Y</math>. Expanding and rearranging, we have <math>VA^{-1} = \left(C^{-1} + VA^{-1}U\right)Y</math>, or <math>\left(C^{-1} + VA^{-1}U\right)^{-1}VA^{-1} = Y</math>. Finally, we substitute into our <math>AX + UY = I</math>, and we have <math>AX + U\left(C^{-1} + VA^{-1}U\right)^{-1}VA^{-1} = I</math>. Thus,
:<math>(A + UCV)^{-1} = X = A^{-1} - A^{-1}U\left(C^{-1} + VA^{-1}U\right)^{-1}VA^{-1}.</math>

We have derived the Woodbury matrix identity.
{{collapse bottom}}

{{collapse top|title=Derivation from LDU decomposition}}
We start by the matrix
:<math>\begin{bmatrix} A & U \\ V & C \end{bmatrix}</math>
By eliminating the entry under the ''A'' (given that ''A'' is invertible) we get

:<math>\begin{bmatrix} I & 0 \\ -VA^{-1} & I \end{bmatrix} 
\begin{bmatrix} A & U \\ V & C \end{bmatrix} = \begin{bmatrix} A & U \\ 0 & C - VA^{-1}U \end{bmatrix}
</math>

Likewise, eliminating the entry above ''C'' gives

:<math>\begin{bmatrix} A & U \\ V & C \end{bmatrix} \begin{bmatrix} I & -A^{-1}U \\ 0 & I \end{bmatrix} 
= \begin{bmatrix} A & 0 \\ V & C-VA^{-1}U \end{bmatrix}
</math>

Now combining the above two, we get

:<math>
  \begin{bmatrix} I & 0 \\ -VA^{-1} & I \end{bmatrix} \begin{bmatrix} A & U \\ V & C \end{bmatrix}\begin{bmatrix} I & -A^{-1}U \\ 0 & I \end{bmatrix} =
  \begin{bmatrix} A & 0 \\ 0 & C - VA^{-1}U \end{bmatrix}
</math>

Moving to the right side gives

:<math>\begin{bmatrix} A & U \\ V & C \end{bmatrix} = \begin{bmatrix} I & 0 \\ VA^{-1} & I \end{bmatrix} \begin{bmatrix} A & 0 \\ 0 & C - VA^{-1}U \end{bmatrix} \begin{bmatrix} I & A^{-1}U \\ 0 & I \end{bmatrix}</math>

which is the LDU decomposition of the block matrix into an upper triangular, diagonal, and lower triangular matrices.

Now inverting both sides gives

: <math>\begin{align}
  \begin{bmatrix} A & U \\ V & C \end{bmatrix}^{-1} 
    &= \begin{bmatrix} I & A^{-1}U \\ 0 & I \end{bmatrix}^{-1} \begin{bmatrix} A & 0 \\ 0 & C - VA^{-1}U \end{bmatrix}^{-1} \begin{bmatrix} I & 0 \\ VA^{-1} & I \end{bmatrix}^{-1} \\[8pt]
    &= \begin{bmatrix} I & -A^{-1}U \\ 0 & I \end{bmatrix} \begin{bmatrix} A^{-1} & 0 \\ 0 & \left(C - VA^{-1}U\right)^{-1} \end{bmatrix} \begin{bmatrix} I & 0 \\ -VA^{-1} & I \end{bmatrix} \\[8pt]
    &= \begin{bmatrix} A^{-1} + A^{-1}U\left(C - VA^{-1}U\right)^{-1}VA^{-1} & -A^{-1}U\left(C - VA^{-1}U\right)^{-1} \\ -\left(C - VA^{-1}U\right)^{-1}VA^{-1} & \left(C - VA^{-1}U\right)^{-1} \end{bmatrix}  \qquad\mathrm{(1)}
\end{align}</math>

We could equally well have done it the other way (provided that ''C'' is invertible) i.e.

: <math>\begin{bmatrix} A & U \\ V & C \end{bmatrix} = \begin{bmatrix} I & UC^{-1} \\ 0 & I \end{bmatrix} \begin{bmatrix} A - UC^{-1}V & 0 \\ 0 & C \end{bmatrix} \begin{bmatrix} I & 0 \\ C^{-1}V  & I\end{bmatrix}</math>

Now again inverting both sides,
: <math>\begin{align}
  \begin{bmatrix} A & U \\ V & C \end{bmatrix}^{-1}
    &= \begin{bmatrix} I & 0 \\ C^{-1}V  & I\end{bmatrix}^{-1} \begin{bmatrix} A - UC^{-1}V & 0 \\ 0 & C \end{bmatrix}^{-1} \begin{bmatrix} I & UC^{-1} \\ 0 & I \end{bmatrix}^{-1} \\[8pt]
    &= \begin{bmatrix} I & 0 \\ -C^{-1}V  & I\end{bmatrix} \begin{bmatrix} \left(A - UC^{-1}V\right)^{-1} & 0 \\ 0 & C^{-1} \end{bmatrix} \begin{bmatrix} I & -UC^{-1} \\ 0 & I \end{bmatrix} \\[8pt]
    &= \begin{bmatrix} \left(A - UC^{-1}V\right)^{-1} & -\left(A - UC^{-1}V\right)^{-1}UC^{-1} \\ -C^{-1}V\left(A - UC^{-1}V\right)^{-1} & C^{-1}V\left(A - UC^{-1}V\right)^{-1}UC^{-1} + C^{-1} \end{bmatrix} \qquad\mathrm{(2)}
\end{align}</math>

Now comparing elements (1, 1) of the RHS of (1) and (2) above gives the Woodbury formula
:<math>\left(A - UC^{-1}V\right)^{-1} = A^{-1} + A^{-1}U\left(C - VA^{-1}U\right)^{-1}VA^{-1}.</math>
{{collapse bottom}}

== Applications ==

This identity is useful in certain numerical computations where ''A''<sup>&minus;1</sup> has already been computed and it is desired to compute (''A''&nbsp;+&nbsp;''UCV'')<sup>&minus;1</sup>.  With the inverse of ''A'' available, it is only necessary to find the inverse of ''C''<sup>−1</sup>&nbsp;+&nbsp;''VA''<sup>−1</sup>''U'' in order to obtain the result using the right-hand side of the identity.  If ''C'' has a much smaller dimension than ''A'', this is more efficient than inverting ''A''&nbsp;+&nbsp;''UCV'' directly. A common case is finding the inverse of a low-rank update ''A''&nbsp;+&nbsp;''UCV'' of ''A'' (where ''U'' only has a few columns and ''V'' only a few rows), or finding an approximation of the inverse of the matrix ''A''&nbsp;+&nbsp;''B'' where the matrix ''B'' can be approximated by a low-rank matrix ''UCV'', for example using the [[singular value decomposition]].

This is applied, e.g., in the [[Kalman filter]] and [[recursive least squares]] methods, to replace the [[parametric solution]], requiring inversion of a state vector sized matrix, with a condition equations based solution. In case of the Kalman filter this matrix has the dimensions of the vector of observations, i.e., as small as 1 in case only one new observation is processed at a time. This significantly speeds up the often real time calculations of the filter.

In the case when ''C'' is the identity matrix ''I'', the matrix <math>I+VA^{-1}U</math> is known in [[numerical linear algebra]] and [[numerical partial differential equations]] as the '''capacitance matrix'''.<ref name="hager"/> 

==See also==
*[[Sherman–Morrison formula]]
*[[Schur complement]]
*[[Matrix determinant lemma]], formula for a rank-''k'' update to a [[determinant]]
*[[Invertible matrix]]
*[[Moore-Penrose pseudoinverse#Updating the pseudoinverse]]

== Notes ==
{{reflist}}
*{{Citation|last1=Press|first1=WH|last2=Teukolsky|first2=SA|last3=Vetterling|first3=WT|last4=Flannery|first4=BP|year=2007|title=Numerical Recipes: The Art of Scientific Computing|edition=3rd|publisher=Cambridge University Press| publication-place=New York|isbn=978-0-521-88068-8|chapter=Section 2.7.3. Woodbury Formula|chapter-url=http://apps.nrbook.com/empanel/index.html?pg=80}}

== External links ==
* [http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/identity.html Some matrix identities]
* {{MathWorld|title=Woodbury formula|urlname=WoodburyFormula}}

[[Category:Linear algebra]]
[[Category:Matrices]]
[[Category:Matrix theory]]
[[Category:Theorems in algebra]]
[[Category:Lemmas]]