Given
an $M\times 1$ vector $\mathbf{a}$ with norm $\| \mathbf{a} \|=\sqrt{M}$
an invertible, Hermitian and Toeplitz matrix $\mathbf{R}$
an $M \times (M-1)$ matrix $\mathbf{C}_n$ whose rank is $M-1$
vector $\mathbf{a}$ is orthogonal to each column of $\mathbf{C}_n$
I'm trying to prove the following equality
$$\frac{1}{M}\left(\mathbf{I} - \mathbf{C}_n \left(\mathbf{C}_n^H\mathbf{R}\mathbf{C}_n \right)^{-1}\mathbf{C}_n^H\mathbf{R} \right) \mathbf{a}=\frac{\mathbf{R}^{-1}\mathbf{a}}{\mathbf{a}^H\mathbf{R}^{-1}\mathbf{a}}$$
$\Pi=I-C_n(C_n^HRC_n)^{-1}C_n^HR$ is an idempotent matrix that maps $C_n$ to zero and leaves $R^{-1}a$ unchanged. Hence it is a rank-one projection onto the linear span of $R^{-1}a$.
Now observe that $R^{-1}a$ and $a^H$ are respectively the right and left eigenvectors of $\Pi$ corresponding to the unit eigenvalue. Therefore $\Pi=cR^{-1}aa^H$ for some scalar $c$. As the rank and trace of a projection matrix coincide, we have $\operatorname{tr}(\Pi)=1$. Hence $c=\frac{1}{a^HR^{-1}a}$ and $$ \Pi=\frac{R^{-1}aa^H}{a^HR^{-1}a}. $$ Consequently, $\frac{1}{M}\Pi a=\frac{R^{-1}a}{a^HR^{-1}a}$.