How do I simplify this matrix equation?

Question

I'm working on a larger proof and am struggling with one portion in which I need get a matrix equation into a desired form. Here is that problem:

I have a matrix equation (left side) that needs to be formatted into another form (right side). I've simplified the left side as much as I could but can't seem to get it to the match the right side. Note that Q and R are invertible. H is not. I've also used MATLAB to verify that the two equations are indeed the same.

How would I go about obtaining this desired form?

Answer 1

I think I got it.

$$ \begin{aligned} (Q &- QH^T(R+HQH^T)^{-1}HQ)H^TR^{-1}) \\ &= ((QH^T(R+HQH^T)^{-1})(QH^T(R+HQH^T)^{-1})^{-1} QH^TR^{-1} \\&\quad -QH^T(R+HQH^T)^{-1}HQH^TR^{-1} ) \\&= (QH^T(R+HQH^T)^{-1})\cdot \underbrace{((QH^T(R+HQH^T)^{-1})^{-1} QH^TR^{-1}-HQH^TR^{-1})}_{=I} \\&= QH^T(R+HQH^T)^{-1} \end{aligned} $$

Where for the last equal sign this is used: \begin{aligned} (&QH^T(R+HQH^T)^{-1})^{-1} QH^TR^{-1}-HQH^TR^{-1} \\&= (R+HQH^T)(QH^T)^{-1}QH^T R^{-1}-HQH^TR^{-1} \\&= (R+HQH^T)R^{-1}-HQH^TR^{-1} \\&= I+HQH^TR^{-1}-HQH^TR^{-1} \\& = I \end{aligned}

Answer 2

Sketch:

Replace $K$ by $L:=Q^{-1}KR$ and re-write the two expressions.

Replace $Q$ by $S:=HQH^{T}$, and re-write the two expressions.

It will suffice to prove $$I-(R+S)^{-1}S=(R+S)^{-1}R$$ and then multiply by $H^{T}$.

Answer 3

This originates from the relationship between block matrix inverse and Schur complements. In general, given a block matrix $$ \pmatrix{A&B\\ C&D} $$ where $A$ and $D$ are nonsingular matrices of probably different sizes, we can write it as a product of block matrices in two different ways: $$ \pmatrix{A&B\\ C&D}= \begin{cases} \pmatrix{I&0\\ CA^{-1}&I}\pmatrix{A&0\\ 0&D-CA^{-1}B}\pmatrix{I&A^{-1}B\\ 0&I},\\ \\ \pmatrix{I&BD^{-1}\\ 0&I}\pmatrix{A-BD^{-1}C&0\\ 0&D}\pmatrix{I&0\\ D^{-1}C&I}. \end{cases} $$ It follows that $$ \pmatrix{X&Y\\ Z&W}:=\pmatrix{A&B\\ C&D}^{-1}= \begin{cases} \pmatrix{I&-A^{-1}B\\ 0&I}\pmatrix{A^{-1}&0\\ 0&(D-CA^{-1}B)^{-1}}\pmatrix{I&0\\ -CA^{-1}&I},\\ \\ \pmatrix{I&0\\ -D^{-1}C&I}\pmatrix{(A-BD^{-1}C)^{-1}&0\\ 0&D^{-1}}\pmatrix{I&-BD^{-1}\\ 0&I}. \end{cases} $$ Therefore, by comparing the two ways of calculating $W$ or $Z$, we get \begin{align} (D-CA^{-1}B)^{-1}&=W=D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1},\tag{1}\\ -(D-CA^{-1}B)^{-1}CA^{-1}&=Z=-D^{-1}C(A-BD^{-1}C)^{-1}.\tag{2} \end{align} Substitute $(1)$ into $(2)$, we obtain $$ \left[D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1}\right]CA^{-1} =D^{-1}C(A-BD^{-1}C)^{-1}. $$ Now, the identity in question follows by putting $A=R,\ B=-H,\ C=H^T$ and $D=Q^{-1}$.

Answer 4

Another way would be as per @ancientmathematician's tip:

$$ K = (Q - Q{H^T}(HQ{H^T} + R)^{-1}HQ){H^T}R^{-1} \\ = Q{H^T}(I - (HQ{H^T} + R)^{-1}HQ{H^T})R^{-1} \\ = Q{H^T}((HQ{H^T} + R)^{-1}R)R^{-1} \\ = Q{H^T}(HQ{H^T} + R)^{-1}, $$

since $(HQ{H^T} + R)^{-1}(HQ{H^T} + R) = I$, so, after rearranging, we get $I - (HQ{H^T} + R)^{-1}HQ{H^T} = (HQ{H^T} + R)^{-1}R,$ which we substitute in the third line of the first chain of equations to get the desired result.