# Is there a way to simplify this inverse?

I have to compute an inverse of the form $$(K + t(XO^T + OX^T))^{-1}$$ where $K,X,O$ are all $n\times n$, and $K$ is symmetric. Assume that $K$ is invertible and that $K^{-1}$ is known.

I would like to simplify this inverse, and if possible write it in terms of powers of $t$. I have looked at using the Woodbury matrix identity. As $XO^T + OX^T$ is symmetric, we can do a Cholesky decomposition to get $XO^T + OX^T = LL^T$. Then:

$$(K + tLL^T)^{-1} = K^{-1} - tK^{-1}L(I + tL^T K^{-1}L)^{-1}L^T K^{-1}$$

From there however I am stuck. Is there a way of simplifying this, or am I chasing a ghost?

Under certain circumstances you can use the Neumann series to write it in powers of $t$. With $M = XO^T + OX^T$ we get
$(K + t M)^{-1} = K^{-1}( I + t M K^{-1})^{-1} = K^{-1} \sum_{k=0}^\infty t^k (- M K^{-1})^k$
which is the expansion in powers of $t$. You can interpret it as analogously to the Taylor expansion in the scalar case. It converges when $M K^{-1}$ has norm $<1$.
If you want an expansion in powers of $t$ this is probably as good as it gets (at least in the direct form it is an infinite series).
• Thanks Andreas, but why is $(K + tM)^{-1} = K^{-1}(I + tK^{-1}M)^{-1})$? And separately, does anything become even simpler if $K = XX^T$? Feb 11, 2015 at 10:27
• Sorry, I confused the order of $K^{-1} M$. I updated the answer. Hmm, I do not see any advantage of knowing the composition of K. Feb 11, 2015 at 17:30
Let:$$(K+tA)^{-1}=(K^{-1}+B_1t+B_2t^2+\cdot\cdot\cdot)$$ Then:$$(K+tA)(K^{-1}+B_1t+B_2t^2+\cdot\cdot\cdot)=I$$ We have:$$(KB_1t+AK^{-1}t+o(t^2)=0$$ $$B=-K^{-1}AK^{-1}$$ $$(K+tA)^{-1}=(K^{-1}-K^{-1}AK^{-1}t+o(t^2))$$ If you keep doing this way, you could get $B_2$, $B_3$, ect.