# An equality in DOA estimation

Suppose we use a uniform linear array for DOA estimation. Assuming neither of the signals and noise are correlated, and the noise power on each sensor is equal, it's well known that the covariance of the received signal can be decomposed as:

$$\boldsymbol{R}=\boldsymbol{A}\boldsymbol{P}\boldsymbol{A}^*+\sigma^2 \boldsymbol{I}=\boldsymbol{S}\boldsymbol{\varLambda_{\mathrm{s}}}\boldsymbol{S}^*+\sigma^2\boldsymbol{G}\boldsymbol{G}^*,$$ where the eigenvectors in $$\boldsymbol{S}$$ and $$\boldsymbol{G}$$ are selected to be orthogonal and normalized.

My question is how to prove the equality $$\boldsymbol{G}\boldsymbol{G}^*=\boldsymbol{I}-\boldsymbol{A}(\boldsymbol{A}^*\boldsymbol{A})^{-1}\boldsymbol{A}^*$$ in the paper Maximum Likelihood Methods for Direction-of-Arrival Estimation. It seems very basic but I can't come up with the proof.

Assume there are $m$ sensors and $d$ sources. As the eigenvectors in $S$ and $G$ of the Hermitian matrix $R$ is selected to be orthogonal and normalized, $$UU^*=\begin{bmatrix}S&G\end{bmatrix}\begin{bmatrix}S^*\\G^*\end{bmatrix}=SS^*+GG^*=UU^{-1}=I_m;$$ $$S^*S=I_d;$$ $$\mathrm{span}\{G\}=\mathrm{span}\{S\}^{\bot}.$$
We know the noise subspace $\mathrm{span}\{G\}$ and $\mathrm{span}\{A\}$ are orthogonal as in the MUSIC algorithm: $$\mathrm{span}\{G\}=\mathrm{span}\{A\}^{\bot}.$$ So $\mathrm{span}\{A\}=\mathrm{span}\{S\}$, which implies their projection matrices into this subspace are equal: $$A(A^*A)^{-1}A^*=S(S^*S)^{-1}S^*.$$ Using $S^*S=I_d$, we get $$A(A^*A)^{-1}A^*=SS^*;$$ with $SS^*+GG^*=I_m$, we get $$GG^*=I_m-SS^*=I_m-A(A^*A)^{-1}A^*.$$