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{{merge|Determinant|discuss=Talk:THIS PAGE#Merger proposal|date=November 2013}}
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In [[mathematics]], in particular [[linear algebra]], the '''matrix determinant lemma'''<ref name="harville">{{cite book | author = Harville, D. A. | year = 1997 | title = Matrix Algebra From a Statistician’s Perspective | publisher = Springer-Verlag | isbn= | doi = }}</ref><ref name="brookes">{{cite web | author = Brookes, M. | title = The Matrix Reference Manual (online) | url = http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/intro.html | year = 2005}}</ref> computes the [[determinant]] of the sum of an [[inverse matrix|invertible]]
[[matrix (mathematics)|matrix]] '''A'''
and the [[dyadic product]], '''u v'''<sup>T</sup>,
of a column [[vector (mathematics)|vector]] '''u''' and a row vector '''v'''<sup>T</sup>. 
 
== Statement ==
Suppose '''A''' is an [[invertible]] [[square matrix]] and '''u''', '''v''' are column [[Vector (geometric)|vectors]]. Then
the matrix determinant lemma states that
:<math>\det(\mathbf{A}+\mathbf{uv}^\mathrm{T}) = (1 + \mathbf{v}^\mathrm{T}\mathbf{A}^{-1}\mathbf{u})\,\det(\mathbf{A}).</math>
 
Here, '''uv'''<sup>T</sup> is the [[outer product]] of two vectors '''u''' and '''v'''.
 
== Proof ==
First the proof of the special case '''A''' = '''I''' follows from the equality:<ref name="ding">{{cite journal |
authors = Ding, J., Zhou, A. | year =  2007 |
title = Eigenvalues of rank-one updated matrices with some applications |
journal = Applied Mathematics Letters | volume = 20 | issue = 12 | pages = 1223–1226 | issn = 0893-9659 |
doi = 10.1016/j.aml.2006.11.016 |
url = http://www.sciencedirect.com/science/article/B6TY9-4N3P02W-5/2/b7f582211325150af4c44674b5e06dd1 }}</ref>
 
:<math>
\begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I}+\mathbf{uv}^\mathrm{T} & \mathbf{u} \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I} & 0 \\ -\mathbf{v}^\mathrm{T} & 1 \end{pmatrix} =
\begin{pmatrix} \mathbf{I} & \mathbf{u} \\ 0 & 1 + \mathbf{v}^\mathrm{T}\mathbf{u} \end{pmatrix}.
</math>
 
The determinant of the left hand side is the product of the determinants of the three matrices.  Since the first and third matrix are triangle matrices with unit diagonal, their determinants are just 1.  The determinant of the middle matrix is our desired value.  The determinant of the right hand side is simply (1 + '''v'''<sup>T</sup>'''u''').  So we have the result:
 
:<math>\det(\mathbf{I}+\mathbf{uv}^\mathrm{T}) = (1 + \mathbf{v}^\mathrm{T}\mathbf{u}).</math>
 
Then the general case can be found as:
:<math>
\begin{align}
\det(\mathbf{A} + \mathbf{uv}^\mathrm{T}) &= \det(\mathbf{A}) \det(\mathbf{I} + \mathbf{A}^{-1}\mathbf{uv}^\mathrm{T})\\
&= \det(\mathbf{A}) (1 + \mathbf{v}^\mathrm{T} \mathbf{A}^{-1}\mathbf{u}).
\end{align}
</math>
 
==Application==
 
If the determinant and inverse of '''A''' are already known, the formula provides a
[[Computationally expensive|numerically cheap]] way
to compute the determinant of '''A''' corrected by the matrix '''uv'''<sup>T</sup>. The computation is relatively cheap because the determinant of '''A'''+'''uv'''<sup>T</sup>
does not have to be computed from scratch (which in general is expensive). Using [[unit vectors]] for '''u''' and/or '''v''', individual columns, rows or elements<ref name="press">{{cite book | authors = William H. Press, Brian P. Flannery, Saul A. Teukolsky, William T. Vetterling | title = [[Numerical Recipes|Numerical Recipes in C: The Art of Scientific Computing]] | pages = 73 | publisher = Cambridge University Press | year = 1992 | isbn = 0-521-43108-5}}</ref> of '''A''' may be manipulated and a correspondingly updated determinant computed relatively cheaply in this way.
 
When the matrix determinant lemma is used in conjunction with the [[Sherman-Morrison formula]], both the inverse and determinant may be conveniently updated together.
 
== Generalization ==
Suppose '''A''' is an [[invertible]] ''n''-by-''n'' matrix and '''U''', '''V''' are ''n''-by-''m'' matrices. Then
:<math>\operatorname{det}(\mathbf{A}+\mathbf{UV}^\mathrm{T}) = \operatorname{det}(\mathbf{I} + \mathbf{V}^\mathrm{T}\mathbf{A}^{-1}\mathbf{U})\operatorname{det}(\mathbf{A}).</math>  
 
In the special case <math>\mathbf{A}=\mathbf{I}</math> this is [[Sylvester's determinant theorem|Sylvester's theorem for determinants]].
 
Given additionally an invertible ''m''-by-''m'' matrix '''W''', the relationship can also be expressed as
:<math>\operatorname{det}(\mathbf{A}+\mathbf{UWV}^\mathrm{T}) = \det(\mathbf{W}^{-1} + \mathbf{V}^\mathrm{T}\mathbf{A}^{-1}\mathbf{U})\det(\mathbf{W})\det(\mathbf{A}).</math>
 
== See also ==
* The [[Sherman-Morrison formula]], which shows how to update the inverse, '''A'''<sup>−1</sup>, to obtain ('''A'''+'''uv'''<sup>T</sup>)<sup>−1</sup>.
* The [[Woodbury matrix identity|Woodbury formula]], which shows how to update the inverse, '''A'''<sup>−1</sup>, to obtain ('''A'''+'''UCV'''<sup>T</sup>)<sup>−1</sup>.
 
== References ==
<references/>
 
[[Category:Linear algebra]]
[[Category:Matrix theory]]
[[Category:Lemmas]]

Revision as of 17:46, 3 March 2014

The author's name is Christy. I've always cherished living in Kentucky but now I'm contemplating other options. For years she's been operating as a travel agent. The favorite hobby for him and his children is to perform lacross and he'll be beginning some thing else along with it.

Here is my blog post - psychic phone readings - pop over to this website -