Notation

Symbol	Meaning
\(\mathbf{H}\), \(\mathbf{x}\)	matrices and vectors are always in bold
\(\mathbf{(\cdot)}^*\)	adjugate matrix
\((\cdot)^H\)	conjugate transpose matrix
\(\det(\cdot)\)	determinant
\(\lvert \cdot \rvert\)	modulus

1. System model

Multiple-input multiple-output (MIMO) system model:

\[\mathbf{y} = \mathbf{H} \mathbf{x} + \mathbf{n}, \qquad \mathbf{H} \in \mathbb{C}^{N_r \times N_t}, \quad \mathbf{n} \sim \mathcal{CN}(\mathbf{0}, \sigma^2 \mathbf{I})\]

We define \(\mathbf{A} = \mathbf{H}^H \mathbf{H} + \sigma^2 \mathbf{I}\). In the \(2 \times 2\), \(3 \times 3\) and \(4 \times 4\) cases below \(\mathbf{H}\) is taken square, \(N_r = N_t = n\), so that \(\mathbf{W}\) is \(n \times n\). The MMSE solution is

\[\hat{\mathbf{x}} = \mathbf{W} \mathbf{y}, \qquad \mathbf{W} = (\mathbf{H}^H \mathbf{H} + \sigma^2 \mathbf{I})^{-1} \mathbf{H}^H = \mathbf{A}^{-1} \mathbf{H}^H\]

The estimator here is explicit. The rest of the post focuses on the derivation of \(\mathbf{A}^{-1}\).

2. \(2 \times 2\) MMSE

We define the matrix \(\mathbf{A}\) in \(2 \times 2\) dimension as

\[\mathbf{A} = \begin{bmatrix} a_{00} & a_{01}\\ a_{01}^H & a_{11} \end{bmatrix}.\]

The adjugate matrix of \(\mathbf{A}\) under \(2 \times 2\) dimension can be obtained by swapping the elements in \(\mathbf{A}\).

\[\mathbf{A}^* = \begin{bmatrix} a_{11} & -a_{01}\\ -a_{01}^H & a_{00} \end{bmatrix}, \qquad\] \[\mathbf{A}^{-1} = \frac{\mathbf{A}^*}{\det(\mathbf{A})}\]

where

\[\det(\mathbf{A}) = a_{00}a_{11} - \lvert a_{01} \rvert^2.\]

3. \(3 \times 3\) MMSE

We define

\[\mathbf{A} = \begin{bmatrix} a_{00} & a_{01} & a_{02}\\ a_{01}^H & a_{11} & a_{12}\\ a_{02}^H & a_{12}^H & a_{22} \end{bmatrix}.\]

Due to the symmetry of \(\mathbf{A}\), only the 6 upper-triangular cofactors are needed:

\[\left\{ \begin{aligned} M_{00}& = a_{11}a_{22} - | a_{12} |^2 \\ M_{11}& = a_{00}a_{22} - | a_{02} |^2 \\ M_{22}& = a_{00}a_{11} - | a_{01} |^2 \\ M_{10}& = a_{02}a_{12}^H - a_{01}a_{22} \\ M_{20}& = a_{01}a_{12} - a_{02}a_{11} \\ M_{21}& = a_{02}a_{01}^H - a_{00}a_{12} \end{aligned} \right.\]

The adjugate matrix of a general \(3 \times 3\) matrix is

\[\mathbf{A}^* = \begin{bmatrix} M_{00} & M_{10} & M_{20}\\[2pt] M_{01} & M_{11} & M_{21}\\[2pt] M_{02} & M_{12} & M_{22} \end{bmatrix} = \begin{bmatrix} a_{11}a_{22} - | a_{12} |^2 & a_{02}a_{12}^H - a_{01}a_{22} & a_{01}a_{12} - a_{02}a_{11}\\[3pt] a_{02}^H a_{12} - a_{01}^H a_{22} & a_{00}a_{22} - | a_{02} |^2 & a_{02}a_{01}^H - a_{00}a_{12}\\[3pt] a_{01}^H a_{12}^H - a_{11}a_{02}^H & a_{01}a_{02}^H - a_{00}a_{12}^H & a_{00}a_{11} - | a_{01} |^2 \end{bmatrix}\]

The determinant is

\[\det(\mathbf{A}) = a_{00}a_{11}a_{22} - a_{00} | a_{12} |^2 - a_{11} | a_{02} |^2 - a_{22} | a_{01} |^2 + a_{01}a_{02}^H a_{12} + a_{01}^H a_{02}a_{12}^H.\]

The inverse is the adjugate over the determinant:

\[\mathbf{A}^{-1} = \frac{\mathbf{A}^*}{\det(\mathbf{A})}.\]

4. \(4 \times 4\) MMSE

If the inverse of a $4 \times 4$ matrix is calculated directly using the adjugate matrix, one must compute 16 third-order algebraic cofactors. Since each cofactor itself requires expansion into a $3 \times 3$ determinant, the computational cost is high. Therefore, we employ a block-based approach, partitioning matrix $\mathbf{A}$ into $2 \times 2$ blocks:

\[\mathbf{A} = \begin{bmatrix} \mathbf{A}_{00} & \mathbf{A}_{01}\\ \mathbf{A}_{01}^H & \mathbf{A}_{11} \end{bmatrix}, \qquad \mathbf{A}_{00} = \begin{bmatrix} a_{00} & a_{01}\\ a_{01}^H & a_{11} \end{bmatrix}, \quad \mathbf{A}_{11} = \begin{bmatrix} a_{22} & a_{23}\\ a_{23}^H & a_{33} \end{bmatrix}, \quad \mathbf{A}_{01} = \begin{bmatrix} a_{02} & a_{03}\\ a_{12} & a_{13} \end{bmatrix}\]

4.1 Method 1: Schur Complement with 2 Divisions

First, eliminate the second block row. The Schur complement of the \(\mathbf{A}_{11}\) block is denoted as \(\mathbf{S}\):

\[\label{eq:S} \mathbf{S} = \mathbf{A}_{00} - \mathbf{A}_{01} \mathbf{A}_{11}^{-1} \mathbf{A}_{01}^H = \mathbf{A}_{00} - \mathbf{C} \mathbf{A}_{01}^H\]

where \(\mathbf{C} = \mathbf{A}_{01} \mathbf{A}_{11}^{-1}\) belongs to the common part of \(\mathbf{S}\) and of the off-diagonal blocks, so the inverse is

\[\mathbf{A}^{-1} = \begin{bmatrix} \mathbf{S}^{-1} & -\mathbf{S}^{-1} \mathbf{C}\\ -\mathbf{C}^H \mathbf{S}^{-1} & \mathbf{A}_{11}^{-1} + \mathbf{C}^H \mathbf{S}^{-1} \mathbf{C} \end{bmatrix}\]

4.2 Method 2: Schur Complement with 1 Division

In the previous section, calculating $\mathbf{S}^{-1}$ and $\mathbf{A}_{11}^{-1}$ requires a total of two divisions. The inverse of a $2 \times 2$ matrix can be written directly using the adjugate matrix. Thus, 2 division operations can be reduced to one.

In equation ($\ref{eq:S}$), the definition of \(\mathbf{S}\) involves \(\mathbf{A}_{11}^{-1}\). To move this inverse matrix to the denominator, one must first express \(\mathbf{S}\) with a common denominator:

\[\begin{aligned} \mathbf{S} &= \mathbf{A}_{00} - \mathbf{A}_{01} \mathbf{A}_{11}^{-1} \mathbf{A}_{01}^H\\ &= \mathbf{A}_{00} - \frac{\mathbf{A}_{01} \mathbf{A}_{11}^* \mathbf{A}_{01}^H}{\det(\mathbf{A}_{11})}\\ &= \frac{\det(\mathbf{A}_{11}) \mathbf{A}_{00} - \mathbf{A}_{01} \mathbf{A}_{11}^* \mathbf{A}_{01}^H}{\det(\mathbf{A}_{11})}\\ &= \frac{\mathbf{T}}{\det(\mathbf{A}_{11})} \end{aligned}\]

\(\mathbf{A}_{01} \mathbf{A}_{11}^*\) is a common part, so we denote it by \(\mathbf{C} = \mathbf{A}_{01} \mathbf{A}_{11}^*\). Then we define

\[\mathbf{T} = \det(\mathbf{A}_{11}) \mathbf{A}_{00} - \mathbf{C} \mathbf{A}_{01}^H\]

There is no division in the expression of \(\mathbf{T}\). The two denominators are folded into

\[D = \det(\mathbf{A}_{11}) \det(\mathbf{T})\]

so that the inverse needs only one reciprocal:

\[\mathbf{A}^{-1} = \frac{1}{D} \begin{bmatrix} \det(\mathbf{A}_{11})^2 \mathbf{T}^* & -\det(\mathbf{A}_{11}) \mathbf{T}^* \mathbf{C}\\ -\det(\mathbf{A}_{11}) \mathbf{C}^H \mathbf{T}^* & \det(\mathbf{T}) \mathbf{A}_{11}^* + \mathbf{C}^H \mathbf{T}^* \mathbf{C} \end{bmatrix}.\]

5. Signal-to-Noise Ratio (SNR)

With \(\hat{\mathbf{x}} = \mathbf{W} \mathbf{y}\) and \(\mathbf{W} = \mathbf{A}^{-1} \mathbf{H}^H\), the error \(\mathbf{e} = \hat{\mathbf{x}} - \mathbf{x}\) splits into a signal part and a noise part,

\[\mathbf{e} = (\mathbf{W} \mathbf{H} - \mathbf{I}) \mathbf{x} + \mathbf{W} \mathbf{n}\]

and the first factor collapses because \(\mathbf{A} = \mathbf{H}^H \mathbf{H} + \sigma^2 \mathbf{I}\):

\[\mathbf{W} \mathbf{H} = \mathbf{A}^{-1} \mathbf{H}^H \mathbf{H} = \mathbf{A}^{-1} (\mathbf{A} - \sigma^2 \mathbf{I}) = \mathbf{I} - \sigma^2 \mathbf{A}^{-1}\]

so

\[\mathbf{e} = -\sigma^2 \mathbf{A}^{-1} \mathbf{x} + \mathbf{A}^{-1} \mathbf{H}^H \mathbf{n}\]

With unit-power symbols and spatially white noise, \(\mathbb{E}[\mathbf{x} \mathbf{x}^H] = \mathbf{I}\) and \(\mathbb{E}[\mathbf{n} \mathbf{n}^H] = \sigma^2 \mathbf{I}\), the error covariance is

\[\mathbf{R}_e = \mathbb{E}[\mathbf{e} \mathbf{e}^H] = \sigma^4 \mathbf{A}^{-2} + \sigma^2 \mathbf{A}^{-1} \mathbf{H}^H \mathbf{H} \mathbf{A}^{-1} = \sigma^2 \mathbf{A}^{-1}\]

because \(\sigma^2 \mathbf{A}^{-1} \mathbf{H}^H \mathbf{H} \mathbf{A}^{-1} = \sigma^2 \mathbf{A}^{-1} - \sigma^4 \mathbf{A}^{-2}\). The mean square error of stream \(i\) is therefore \(\sigma^2 [\mathbf{A}^{-1}]_{ii}\). The signal-to-noise ratio of stream \(i\) is

\[\mathrm{SNR}_i = \frac{1 - \sigma^2 [\mathbf{A}^{-1}]_{ii}}{\sigma^2 [\mathbf{A}^{-1}]_{ii}} = \frac{1}{\sigma^2 [\mathbf{A}^{-1}]_{ii}} - 1\]

The inverse matrix is equal to the adjugate matrix divided by the determinant. Therefore, its diagonal elements are derived from the algebraic cofactors constructed using the aforementioned method:

\[\mathbf{A}^{-1} = \frac{\mathbf{A}^*}{\det(\mathbf{A})}, \qquad [\mathbf{A}^{-1}]_{ii} = \frac{M_{ii}}{\det(\mathbf{A})}\]

The block decomposition method for a \(4 \times 4\) matrix constructs a Schur complement, so cofactors do not appear, and the diagonal entries must be completed using block matrix inverses.

\[\mathbf{A}^{-1} = \begin{bmatrix} \mathbf{S}^{-1} & \cdot \\ \cdot & \mathbf{A}_{11}^{-1} + \mathbf{A}_{11}^{-1} \mathbf{A}_{01}^H \mathbf{S}^{-1} \mathbf{A}_{01} \mathbf{A}_{11}^{-1} \end{bmatrix}\]