I'm working through the exercises in my linear algebra book. Normally they show the answers to odd-numbered exercises, however for some reason not for these, so I have no idea what to do.
They want us to verify by block multiplication that the inverse of a matrix, if partitioned as shown, is as claimed (assume that all inverses exist as needed).
Then after this, partition the matrix below, so that you can apply the formula acquired from the above exercise to calculate the inverse.
I honestly have no idea where to start, because there aren't any examples in the book showing how it's done. I found someone else asking a similar question here, but I still don't understand how to solve it.
Best Answer
In your problem you are given a candidate for the inverse. All you need to do is to verify that \begin{equation} \begin{pmatrix} I & B \\ C & I \end{pmatrix} \begin{pmatrix} (I - BC)^{-1} & -(I - BC)^{-1} B \\ -C (I-BC)^{-1} & I + C(I-BC)^{-1} B \end{pmatrix} = \begin{pmatrix} I & 0\\ 0 & I \end{pmatrix}. \end{equation} There are four blocks to compute. I will do the block $(2,2)$ and leave the three others to you. We have \begin{equation} C(-(I-BC)^{-1}B) + I(I+C(I-BC)^{-1}B) = I - C(I-BC)^{-1}B + C(I-BC)^{-1}B = I. \end{equation} which is exactly what we want.
It can be that you are missing the following piece of the puzzle. Suppose $A$ and $B$ are matrices such that the product $AB$ is defined and that the two matrices are partitioned conformally, then \begin{equation} \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} = \begin{pmatrix} A_{11}B_{11} + A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\ A_{21}B_{11} + A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{pmatrix} \end{equation} The term "conformally" merely means that all the product on the right hand side are defined.
The formula itself is the natural extension of how you multiply two matrices square of dimension 2. In reality, it is the just the statement that any inner product $S = x^T y$ can be done gradually, i.e. if $x = (x_1, x_2)^T$ and $y = (y_1,y_2)^T$ are vectors which have been partitioned conformally, i.e. such that $x_1$ and $y_1$ have the same length and similarly for $x_2$ and $y_2$, then \begin{equation} s = (x_1^Ty_1) + (x_2^T y_2). \end{equation}
I hope this helps.