This article is a draft and as such there might be typos and other inaccuracies.
In this article we’ll derive the matrix inversion lemma, also known as the Sherman-Morrisson-Woodbury formula. At first it might seem like a very boring piece of linear algebra, but it has a few nifty uses, as we’ll see in one of the followup articles.
Let’s start with the following block matrix:
M=[AVUB]
We’ll do an LDU decomposition in two different ways, which basically direclty gives us the end formula. Eliminating the bototm left element we get the following:
[I−VA−10I][AVUB]=[A0UB−VA−1U]
The $B - V A^{-1} U$ is called a Schur complement and is generally defined as follows:
M/A:=VA−1U
We’ll use this notation later to make things easier to read. Moving on with the decomposition, we’ll now eliminate U.
[A0UB−VA−1U][I0−A−1UI]=[A00B−VA−1U]
Putting the two equations above together we get the following:
That’s it for the first part, now we’ll do the same, but eliminating the top-right element from the left first.
[I0−UB−1I][AVUB]=[A−UB−1VV0B]
Here we get the other Schur complement, which we’ll note as M/B=A−UB−1V. We can substitute it in straight away this time.
[M/BV0B][I−B−1V0I]=[M/B00B]
As before, we’ll write it out as a single equation:
[I0−UB−1I][AVUB][I−B−1V0I]=[M/B00B]
Now we express the matrix M in terms of the other two (notice the newly added inverse signs):
[AVUB]=[I0−UB−1I]−1[M/B00B][I−B−1V0I]−1
Lastly, we just take the inverse of both sides:
[AVUB]−1=([I0−UB−1I]−1[M/B00B][I−B−1V0I]−1)−1=[I−B−1V0I][M/B00B]−1[I0−UB−1I]=[I−B−1V0I][(M/B)−100B−1][I0−UB−1I]=[(M/B)−1−B−1V(M/B)−10B−1][I0−UB−1I]=[(M/B)−1−B−1V(M/B)−1−(M/B)−1UB−1B−1V(M/B)−1UB−1+B−1]=[(A−UB−1V)−1−B−1V(A−UB−1V)−1−(A−UB−1V)−1UB−1B−1V(A−UB−1V)−1UB−1+B−1]notice the inverses cancelling out
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