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| {{merge|Determinant|discuss=Talk:THIS PAGE#Merger proposal|date=November 2013}}
| | Andrew Simcox is the name his parents gave him and he completely loves this name. To play lacross is the factor I love most of all. Distributing production is how he makes a living. Ohio is exactly where my house is but my husband desires us to move.<br><br>Feel free to visit my site ... psychic readers; [http://gcjcteam.org/index.php?mid=etc_video&document_srl=696611&sort_index=regdate&order_type=desc More about the author], |
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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]]
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| [[matrix (mathematics)|matrix]] '''A'''
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| and the [[dyadic product]], '''u v'''<sup>T</sup>,
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| of a column [[vector (mathematics)|vector]] '''u''' and a row vector '''v'''<sup>T</sup>.
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| == Statement ==
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| Suppose '''A''' is an [[invertible]] [[square matrix]] and '''u''', '''v''' are column [[Vector (geometric)|vectors]]. Then
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| the matrix determinant lemma states that
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| :<math>\det(\mathbf{A}+\mathbf{uv}^\mathrm{T}) = (1 + \mathbf{v}^\mathrm{T}\mathbf{A}^{-1}\mathbf{u})\,\det(\mathbf{A}).</math>
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| Here, '''uv'''<sup>T</sup> is the [[outer product]] of two vectors '''u''' and '''v'''.
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| == Proof ==
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| First the proof of the special case '''A''' = '''I''' follows from the equality:<ref name="ding">{{cite journal |
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| authors = Ding, J., Zhou, A. | year = 2007 |
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| title = Eigenvalues of rank-one updated matrices with some applications |
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| journal = Applied Mathematics Letters | volume = 20 | issue = 12 | pages = 1223–1226 | issn = 0893-9659 |
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| doi = 10.1016/j.aml.2006.11.016 |
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| url = http://www.sciencedirect.com/science/article/B6TY9-4N3P02W-5/2/b7f582211325150af4c44674b5e06dd1 }}</ref>
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| :<math>
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| \begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
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| \begin{pmatrix} \mathbf{I}+\mathbf{uv}^\mathrm{T} & \mathbf{u} \\ 0 & 1 \end{pmatrix}
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| \begin{pmatrix} \mathbf{I} & 0 \\ -\mathbf{v}^\mathrm{T} & 1 \end{pmatrix} =
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| \begin{pmatrix} \mathbf{I} & \mathbf{u} \\ 0 & 1 + \mathbf{v}^\mathrm{T}\mathbf{u} \end{pmatrix}.
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| </math>
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| 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:
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| :<math>\det(\mathbf{I}+\mathbf{uv}^\mathrm{T}) = (1 + \mathbf{v}^\mathrm{T}\mathbf{u}).</math>
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| Then the general case can be found as:
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| :<math>
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| \begin{align}
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| \det(\mathbf{A} + \mathbf{uv}^\mathrm{T}) &= \det(\mathbf{A}) \det(\mathbf{I} + \mathbf{A}^{-1}\mathbf{uv}^\mathrm{T})\\
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| &= \det(\mathbf{A}) (1 + \mathbf{v}^\mathrm{T} \mathbf{A}^{-1}\mathbf{u}).
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| \end{align}
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| </math>
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| ==Application==
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| If the determinant and inverse of '''A''' are already known, the formula provides a
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| [[Computationally expensive|numerically cheap]] way
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| 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.
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| When the matrix determinant lemma is used in conjunction with the [[Sherman-Morrison formula]], both the inverse and determinant may be conveniently updated together.
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| == Generalization ==
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| Suppose '''A''' is an [[invertible]] ''n''-by-''n'' matrix and '''U''', '''V''' are ''n''-by-''m'' matrices. Then
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| :<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>
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| In the special case <math>\mathbf{A}=\mathbf{I}</math> this is [[Sylvester's determinant theorem|Sylvester's theorem for determinants]].
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| Given additionally an invertible ''m''-by-''m'' matrix '''W''', the relationship can also be expressed as
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| :<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>
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| == 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>.
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| * 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>.
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| == References ==
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| <references/>
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| [[Category:Linear algebra]]
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| [[Category:Matrix theory]]
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| [[Category:Lemmas]]
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Andrew Simcox is the name his parents gave him and he completely loves this name. To play lacross is the factor I love most of all. Distributing production is how he makes a living. Ohio is exactly where my house is but my husband desires us to move.
Feel free to visit my site ... psychic readers; More about the author,