# proof of inverse of matrix with small-rank adjustment

We will first prove the formula when $A=I$.

Suppose that $R^{-1}+Y^{T}X$ is invertible  . Thus

 $(R^{-1}+Y^{T}X)(R^{-1}+Y^{T}X)^{-1}=I.$

and

 $R^{-1}(R^{-1}+Y^{T}X)^{-1}+Y^{T}X(R^{-1}+Y^{T}X)^{-1}=I.$

Multiply by $XR$ from the left, and multiply by $Y^{T}$ from the right, we get

 $X(R^{-1}+Y^{T}X)^{-1}Y^{T}+XRY^{T}X(R^{-1}+Y^{T}X)^{-1}Y^{T}=XRY^{T}.$

The right hand side is equal to $B-I$, while the left hand side can be factorized as

 $(I+XRY^{T})X(R^{-1}+Y^{T}X)^{-1}Y^{T}.$

So,

 $B\cdot(X(R^{-1}+Y^{T}X)^{-1}Y^{T})=B-I.$

After rearranging, we obtain

 $I=B(I-X(R^{-1}+Y^{T}X)^{-1}Y^{T}).$

Therefore

 $(I+XRY^{T})^{-1}=I-X(R^{-1}+Y^{T}X)^{-1}Y^{T}{}$ (*)

For the general case $B=A+XRY^{T}$, consider

 $BA^{-1}=I+XRY^{T}A^{-1}.$

We can apply (*) with $Y^{T}$ replaced by $Y^{T}A^{-1}$.

Title proof of inverse of matrix with small-rank adjustment ProofOfInverseOfMatrixWithSmallrankAdjustment 2013-03-22 15:46:08 2013-03-22 15:46:08 kshum (5987) kshum (5987) 4 kshum (5987) Proof msc 15A09