I found the algebra identity in my math textbook:

\begin{equation} \text{If } K'X = 0 \text{ for } K \text{ of full column rank and } H \text{ is positive definite, then:} \end{equation} \begin{equation} K(K'HK)^{-1}K' = H^{-1} - H^{-1}X (X'H^{-1}X)^{-1} X'H^{-1} = P \end{equation}

Can someone please tell me - is this a version of the Woodbury Matrix Identity?




1 Answer 1


Presumably $K'$ means the transpose or conjugate transpose of $K$ (depending on whether the underlying field is $\mathbb R$ or $\mathbb C$). Let $H$ be $n\times n$. The equality in question is true only if $\operatorname{rank}(K)+\operatorname{rank}(X)=n$. Otherwise, it is false in general. A counterexample is given by $n=3$, $H=I_3$, $K=(1,0,0)'$ and $X=(0,1,0)'$.

When the aforementioned rank condition is satisfied, we have $\operatorname{rank}(Z)+\operatorname{rank}(Y)=n$ and $Z'Y=0$ where $Z=H^{1/2}K$ and $Y=H^{-1/2}X$. Therefore the column spaces of $Z$ and $Y$ are orthogonal complement of $\mathbb R^n$ or $\mathbb C^n$. It follows that the orthogonal projections onto these two column spaces must sum to $I_n$, i.e., $$ Z(Z'Z)^{-1}Z'=I_n-Y(Y'Y)^{-1}Y'. $$ Consequently, $H^{-1/2}Z(Z'Z)^{-1}Z'H^{-1/2}=H^{-1}-H^{-1/2}Y(Y'Y)^{-1}Y'H^{-1/2}$ and the result follows.


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