Shallow water equations

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

$\left(\mathbf {A} +\mathbf {UBV} \right)^{-1}=\mathbf {A} ^{-1}-\mathbf {A} ^{-1}\mathbf {UB} \left(\mathbf {B} +\mathbf {BVA} ^{-1}\mathbf {UB} \right)^{-1}\mathbf {BVA} ^{-1}$

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

$\left(\mathbf {A} +\mathbf {UBV} \right)^{-1}=\mathbf {A} ^{-1}-\mathbf {A} ^{-1}\mathbf {U} \left(\mathbf {B} ^{-1}+\mathbf {VA} ^{-1}\mathbf {U} \right)^{-1}\mathbf {VA} ^{-1}.$

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

Verification

First notice that

$\left(\mathbf {A} +\mathbf {UBV} \right)\mathbf {A} ^{-1}\mathbf {UB} =\mathbf {UB} +\mathbf {UBVA} ^{-1}\mathbf {UB} =\mathbf {U} \left(\mathbf {B} +\mathbf {BVA} ^{-1}\mathbf {UB} \right).$

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

$\left(\mathbf {A} +\mathbf {UBV} \right)\left(\mathbf {A} ^{-1}-\mathbf {A} ^{-1}\mathbf {UB} \left(\mathbf {B} +\mathbf {BVA} ^{-1}\mathbf {UB} \right)^{-1}\mathbf {BVA} ^{-1}\right)$

$=\mathbf {I} _{p}+\mathbf {UBVA} ^{-1}-\mathbf {U} \left(\mathbf {B} +\mathbf {BVA} ^{-1}\mathbf {UB} \right)\left(\mathbf {B} +\mathbf {BVA} ^{-1}\mathbf {UB} \right)^{-1}\mathbf {BVA} ^{-1}$

$=\mathbf {I} _{p}+\mathbf {UBVA} ^{-1}-\mathbf {UBVA} ^{-1}=\mathbf {I} _{p}\!$

which verifies that it is the inverse.

So we get that—if A⁻¹ and $\left(\mathbf {B} +\mathbf {BVA} ^{-1}\mathbf {UB} \right)^{-1}$ exist, then $\left(\mathbf {A} +\mathbf {UBV} \right)^{-1}$ exists and is given by the theorem above.

Special cases

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

$\left(\mathbf {A} +\mathbf {B} \right)^{-1}=\mathbf {A} ^{-1}-\mathbf {A} ^{-1}\mathbf {B} \left(\mathbf {B} +\mathbf {BA} ^{-1}\mathbf {B} \right)^{-1}\mathbf {BA} ^{-1}.$

Remembering the identity

$\left(\mathbf {A} \mathbf {B} \right)^{-1}=\mathbf {B} ^{-1}\mathbf {A} ^{-1}.$

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

$\left(\mathbf {A} +\mathbf {B} \right)^{-1}=\mathbf {A} ^{-1}-\mathbf {A} ^{-1}\left(\mathbf {I} +\mathbf {B} \mathbf {A} ^{-1}\right)^{-1}\mathbf {B} \mathbf {A} ^{-1}.$

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

$\left(\mathbf {A} +\mathbf {uv} ^{\mathrm {T} }\right)^{-1}=\mathbf {A} ^{-1}-{\frac {\mathbf {A} ^{-1}\mathbf {uv} ^{\mathrm {T} }\mathbf {A} ^{-1}}{1+\mathbf {v} ^{\mathrm {T} }\mathbf {A} ^{-1}\mathbf {u} }}.$

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+uv^T quickly.

If we set A = I_p and B = I_q, we get

$\left(\mathbf {I} _{p}+\mathbf {UV} \right)^{-1}=\mathbf {I} _{p}-\mathbf {U} \left(\mathbf {I} _{q}+\mathbf {VU} \right)^{-1}\mathbf {V} .$

In particular, if q = 1, then

$\left(\mathbf {I} +\mathbf {uv} ^{\mathrm {T} }\right)^{-1}=\mathbf {I} -{\frac {\mathbf {uv} ^{\mathrm {T} }}{1+\mathbf {v} ^{\mathrm {T} }\mathbf {u} }}.$

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, 21 year-old Glazier James Grippo from Edam, enjoys hang gliding, industrial property developers in singapore developers in singapore and camping. Finds the entire world an motivating place we have spent 4 months at Alejandro de Humboldt National Park. preprint Template:MR
• Moore-Penrose pseudoinverse#Updating the pseudoinverse