# why does the blockwise inversion formula work?

I used this http://en.wikipedia.org/wiki/Matrix_inverse#Blockwise_inversion formula to get the inverse of a partitioned matrix, and it works great. What I don't understand is why exactly it works.

If I have a matrix

$$\left(\begin{array}{cc}A& B\\C& D\end{array}\right)$$

and its inverse $$\left(\begin{array}{cc}W& X\\Y& Z\end{array}\right)$$

I can see that AW + BY = I, CX + DZ = I, and the other products are zero matrices, but when I try to use these relationships to build the blockwise formula back up, I don't get it right. Can anyone prove this formula to me?

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Isn't it the other way around with AW + BY and CX + DZ being identity matrices? –  Wonder Apr 25 '12 at 13:11
yep, sorry bout that –  user29373 Apr 25 '12 at 13:39

## 1 Answer

Since we already know the form of the solution, it's not too hard to just dive in and steer our way towards the result. If $$\begin{bmatrix}A & B \\ C & D\end{bmatrix}\begin{bmatrix}W & X \\ Y & Z\end{bmatrix} = \begin{bmatrix}I & 0 \\ 0 & I\end{bmatrix},$$ then $$AX + BZ = 0.$$ Assuming $A$ is invertible, $$X + A^{-1}BZ = 0,$$ $$X = -A^{-1}BZ.$$ Now $$CX + DZ = I,$$ so $$-CA^{-1}BZ + DZ = I,$$ $$(D - CA^{-1}B)Z = I,$$ $$Z = (D - CA^{-1}B)^{-1}$$ as long as $D - CA^{-1}B$ is also invertible. Now that we have one of the entries of the blockwise inverse, we can start substituting it into the other products and simplifying them. Do you think you can take it from here? You'll also need to consider that $$\begin{bmatrix}W & X \\ Y & Z\end{bmatrix}\begin{bmatrix}A & B \\ C & D\end{bmatrix} = \begin{bmatrix}I & 0 \\ 0 & I\end{bmatrix}.$$

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Because in the other products there are only $W$ and $Y$ (not $X$ and $Z$ that are now determined) I am asking myself: is not necessary some other hypothesis apart from the invertibility of $A$? –  Giuseppe Tortorella Apr 25 '12 at 13:21
thanks! - I was just making a dumb arithmetical mistake as per usual –  user29373 Apr 25 '12 at 13:39
@Giuseppe: Good question. We also need to consider that the product with $\begin{bmatrix}A & B \\ C & D\end{bmatrix}$ and $\begin{bmatrix}W & X \\ Y & Z\end{bmatrix}$ interchanged is also the identity. –  Rahul Apr 25 '12 at 14:34