How to Prove $(GG^H +I_M)^{-1} g_k = \frac {R_k^{-1} g_k}{g^h_kR_k^{-1}g_k+1}$

Question

where
 G is MXK matrix,  g_k is kth colum of G, I_M is MXM identity matrix

  and H denotes the conjugate transpose (hermitian operator)

and $R_k= \sum_{i=1, i\neq k}^{K}g_ig_i^H +I_M$

and R_k is same as G except kth column removed.

Note: It is used in MMSE estimation for MIMO and reference said that it require Woodbury matrix identityenter image description here to prove.

Answer 1

Using block matrix multiplication we can write $GG^H$ as

$$ GG^H + I_M= \begin{bmatrix} g_1\ \cdots \ g_k \end{bmatrix} \begin{bmatrix} g_1^H\\ \vdots\\ g_k^H \end{bmatrix} + I_M = \sum_{i=1}^{K} g_i g_i^H + I_M = R_k + g_k \cdot g_k^H$$

Now, we can apply the Woodbury matrix identity using $A = R_k$, $U = g_k$, $C=1$ and $V=g_k^H$

$$\left( R_k + g_k \cdot g_k^H \right)^{-1} = R_k^{-1} - R_k^{-1}g_k \left(1 + g_k^H R_k^{-1} g_k \right)^{-1}g_k^H R_k^{-1} = R_k^{-1} - \frac{R_k^{-1}g_kg_k^H R_k^{-1}}{1 + g_k^H R_k^{-1} g_k} $$

Finally, multiplying by $g_k$ we reach the result

$$ \begin{align*} \left( GG^H + I_M \right)^{-1}g_k &= \left( R_k + g_k \cdot g_k^H \right)^{-1}g_k = R_k^{-1}g_k - \frac{R_k^{-1}g_kg_k^H R_k^{-1}g_k}{1 + g_k^H R_k^{-1} g_k} \\ &= \frac{R_k^{-1}g_k + R_k^{-1}g_kg_k^H R_k^{-1} g_k - R_k^{-1}g_kg_k^H R_k^{-1} g_k}{1 + g_k^H R_k^{-1} g_k}\\ &= \frac{R_k^{-1}g_k}{1 + g_k^H R_k^{-1} g_k} \end{align*} $$