Binomial inverse theorem

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. Nonsingularity of the latter requires that B⁻¹ exist since it equals B(I+VA^—1UB) and the rank of the latter cannot exceed the rank of B.

Since B is invertible, the two B terms flanking the parenthetical 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 Woodbury matrix identity, which can also be derived using matrix blockwise inversion.

A more general formula exists when B is singular and possibly even non-square:

$(\mathbf{A+UBV})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1}\mathbf{U}(\mathbf{I+BVA}^{-1}\mathbf{U})^{-1}\mathbf{BVA}^{-1}.$

Formulas also exist for certain cases in which A is singular.

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:

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

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

$= \mathbf{I}_p + \mathbf{UBVA}^{-1} - \mathbf{U BVA}^{-1} = \mathbf{I}_p \!$

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

First

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

$\left(\mathbf{C} \mathbf{D}\right)^{-1} = \mathbf{D}^{-1} \mathbf{C}^{-1} ,$

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

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

Second

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 the Sherman-Morrison formula:

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

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 for various u and v.

Third

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

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

which is a particular case of the Sherman-Morrison formula given above.

See also

• Invertible matrix
• Matrix determinant lemma
• Moore-Penrose pseudoinverse#Updating the pseudoinverse