Shallow water equations: Difference between revisions

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Conservative form: The equation quoted was missing part of the pressure term, which I've put on the right hand side. (Jody Klymak) jklymak@gmail.com
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In [[mathematics]], the '''Binomial Inverse Theorem''' is useful for expressing [[matrix (mathematics)|matrix]] inverses in different ways.
 
If '''A''', '''U''', '''B''', '''V''' are matrices of sizes ''p''×''p'', ''p''×''q'', ''q''×''q'', ''q''×''p'', respectively, then
 
:<math>
\left(\mathbf{A}+\mathbf{UBV}\right)^{-1}=
\mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{UB}\left(\mathbf{B}+\mathbf{BVA}^{-1}\mathbf{UB}\right)^{-1}\mathbf{BVA}^{-1}
</math>
 
provided '''A''' and '''B''' + '''BVA'''<sup>−1</sup>'''UB''' are nonsingular. Note that if '''B''' is invertible, the two '''B''' terms flanking the quantity inverse in the right-hand side can be replaced with ('''B'''<sup>−1</sup>)<sup>−1</sup>, which results in
 
:<math>
\left(\mathbf{A}+\mathbf{UBV}\right)^{-1}=
\mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{U}\left(\mathbf{B}^{-1}+\mathbf{VA}^{-1}\mathbf{U}\right)^{-1}\mathbf{VA}^{-1}.
</math>
 
This is the [[matrix inversion lemma]], which can also be derived using [[Invertible matrix#Blockwise inversion|matrix blockwise inversion]].
 
==Verification==
First notice that
:<math>\left(\mathbf{A} + \mathbf{UBV}\right) \mathbf{A}^{-1}\mathbf{UB} = \mathbf{UB} + \mathbf{UBVA}^{-1}\mathbf{UB} = \mathbf{U} \left(\mathbf{B} + \mathbf{BVA}^{-1}\mathbf{UB}\right).</math>
 
Now multiply the matrix we wish to invert by its alleged inverse
:<math>\left(\mathbf{A} + \mathbf{UBV}\right) \left( \mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{UB}\left(\mathbf{B} + \mathbf{BVA}^{-1}\mathbf{UB}\right)^{-1}\mathbf{BVA}^{-1} \right) </math>
:<math>= \mathbf{I}_p + \mathbf{UBVA}^{-1} - \mathbf{U} \left(\mathbf{B} + \mathbf{BVA}^{-1}\mathbf{UB}\right) \left(\mathbf{B} + \mathbf{BVA}^{-1}\mathbf{UB}\right)^{-1}\mathbf{BVA}^{-1} </math>
:<math>= \mathbf{I}_p + \mathbf{UBVA}^{-1} - \mathbf{U BVA}^{-1} = \mathbf{I}_p \!</math>
 
which verifies that it is the inverse.
 
So we get that—if '''A'''<sup>−1</sup> and <math>\left(\mathbf{B} + \mathbf{BVA}^{-1}\mathbf{UB}\right)^{-1}</math> exist, then <math>\left(\mathbf{A} + \mathbf{UBV}\right)^{-1}</math> exists and is given by the theorem above.<ref name="strang">{{cite book | author = Gilbert Strang | title = Introduction to Linear Algebra | edition = 3rd edition | year = 2003 | publisher = Wellesley-Cambridge Press: Wellesley, MA | isbn = 0-9614088-9-8}}</ref>
 
==Special cases==
If ''p'' = ''q'' and '''U''' = '''V''' = '''I'''<sub>''p''</sub> is the identity matrix, then
 
:<math>
\left(\mathbf{A}+\mathbf{B}\right)^{-1} = \mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{B}\left(\mathbf{B}+\mathbf{BA}^{-1}\mathbf{B}\right)^{-1}\mathbf{BA}^{-1}.
</math>
 
Remembering the identity
:<math>
\left(\mathbf{A} \mathbf{B}\right)^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1} .
</math>
we can also express the previous equation in the simpler form as
 
:<math>
\left(\mathbf{A}+\mathbf{B}\right)^{-1} = \mathbf{A}^{-1} - \mathbf{A}^{-1}\left(\mathbf{I}+\mathbf{B}\mathbf{A}^{-1}\right)^{-1}\mathbf{B}\mathbf{A}^{-1}.
</math>
 
If '''B''' = '''I'''<sub>''q''</sub> is the identity matrix and ''q'' = 1, then '''U''' is a column vector, written '''u''', and '''V''' is a row vector, written '''v'''<sup>T</sup>.  Then the theorem implies
 
:<math>
\left(\mathbf{A}+\mathbf{uv}^\mathrm{T}\right)^{-1} = \mathbf{A}^{-1}- \frac{\mathbf{A}^{-1}\mathbf{uv}^\mathrm{T}\mathbf{A}^{-1}}{1+\mathbf{v}^\mathrm{T}\mathbf{A}^{-1}\mathbf{u}}.
</math>
 
This is useful if one has a matrix <math>A</math> with a known inverse '''A'''<sup>−1</sup> and one needs to invert matrices of the form '''A'''+'''uv'''<sup>T</sup> quickly.
 
If we set '''A''' = '''I'''<sub>''p''</sub> and '''B''' = '''I'''<sub>''q''</sub>, we get
:<math>\left(\mathbf{I}_p + \mathbf{UV}\right)^{-1} = \mathbf{I}_p - \mathbf{U}\left(\mathbf{I}_q + \mathbf{VU}\right)^{-1}\mathbf{V}.</math>
 
In particular, if ''q'' = 1, then
 
:<math>\left(\mathbf{I}+\mathbf{uv}^\mathrm{T}\right)^{-1} = \mathbf{I} - \frac{\mathbf{uv}^\mathrm{T}}{1+\mathbf{v}^\mathrm{T}\mathbf{u}}.</math>
 
==See also==
*[[Woodbury matrix identity]]
*[[Sherman-Morrison formula]]
*[[Invertible matrix]]
*[[Matrix determinant lemma]]
* For certain cases where ''A'' is singular and also [[Moore-Penrose pseudoinverse]], see 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}}
* [[Moore-Penrose pseudoinverse#Updating the pseudoinverse]]
 
==References==
<references/>
 
[[Category:Linear algebra]]
[[Category:Matrix theory]]
[[Category:Theorems in algebra]]

Revision as of 22:54, 28 October 2013

In mathematics, the Binomial Inverse Theorem is useful for expressing matrix inverses in different ways.

If A, U, B, V are matrices of sizes p×p, p×q, q×q, q×p, respectively, then

(A+UBV)1=A1A1UB(B+BVA1UB)1BVA1

provided A and B + BVA−1UB are nonsingular. Note that if B is invertible, the two B terms flanking the quantity inverse in the right-hand side can be replaced with (B−1)−1, which results in

(A+UBV)1=A1A1U(B1+VA1U)1VA1.

This is the matrix inversion lemma, which can also be derived using matrix blockwise inversion.

Verification

First notice that

(A+UBV)A1UB=UB+UBVA1UB=U(B+BVA1UB).

Now multiply the matrix we wish to invert by its alleged inverse

(A+UBV)(A1A1UB(B+BVA1UB)1BVA1)
=Ip+UBVA1U(B+BVA1UB)(B+BVA1UB)1BVA1
=Ip+UBVA1UBVA1=Ip

which verifies that it is the inverse.

So we get that—if A−1 and (B+BVA1UB)1 exist, then (A+UBV)1 exists and is given by the theorem above.[1]

Special cases

If p = q and U = V = Ip is the identity matrix, then

(A+B)1=A1A1B(B+BA1B)1BA1.

Remembering the identity

(AB)1=B1A1.

we can also express the previous equation in the simpler form as

(A+B)1=A1A1(I+BA1)1BA1.

If B = Iq is the identity matrix and q = 1, then U is a column vector, written u, and V is a row vector, written vT. Then the theorem implies

(A+uvT)1=A1A1uvTA11+vTA1u.

This is useful if one has a matrix A with a known inverse A−1 and one needs to invert matrices of the form A+uvT quickly.

If we set A = Ip and B = Iq, we get

(Ip+UV)1=IpU(Iq+VU)1V.

In particular, if q = 1, then

(I+uvT)1=IuvT1+vTu.

See also

References

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