# Is this the Woodbury Matrix Identity?

I found the algebra identity in my math textbook:

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

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

Thanks!

References:

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.