Binomial inverse theorem

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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.

1 Verification
2 Special cases
3 See also
4 References

[edit] 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.^[1]

[edit] Special cases

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

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

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

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