Notation
Symbol Meaning \(\mathbf{H}\), \(\mathbf{x}\) matrices and vectors are always in bold \(a_{ij}\), \(l_{ij}\) scalars are in italic with subscripts \(d_i\) real positive pivot of the LDL decomposition \((\cdot)^H\) conjugate transpose \(\det(\cdot)\) determinant \(\lvert \cdot \rvert\) modulus \(\operatorname{rsqrt}(\cdot)\) reciprocal square root, defined by \(\operatorname{rsqrt}(\cdot) = 1 / \sqrt{\cdot}\), counted as one operation of its own \(\operatorname{diag}(\cdot)\) diagonal matrix built from the listed entries \(\mathbf{L}\) lower triangular factor; its diagonal is unit for the LDL decomposition \(\mathbf{D}\) real positive diagonal factor \(\hat{\mathbf{z}}\) forward substitution intermediate
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 \(\mathbf{A} = \mathbf{H}^H \mathbf{H} + \sigma^2 \mathbf{I}\) and \(\mathbf{z} = \mathbf{H}^H \mathbf{y}\). The MMSE solution is
\[\hat{\mathbf{x}} = (\mathbf{H}^H \mathbf{H} + \sigma^2 \mathbf{I})^{-1} \mathbf{H}^H \mathbf{y} = \mathbf{A}^{-1} \mathbf{z},\]so the estimate is the solution of the linear equation
\[\mathbf{A} \hat{\mathbf{x}} = \mathbf{z}.\]The estimation approach does not construct weight matrix \(\mathbf{W}\) and is therefore referred to as implicit estimation. The matrix \(\mathbf{A}\) is Hermitian, and with \(\sigma^2 > 0\) it is also positive definite even when \(\mathbf{H}\) is rank-deficient. This is the property both decompositions below rely on: it guarantees that the triangular factors exist.
2. Cholesky decomposition
A Hermitian positive definite matrix \(\mathbf{A}\) can be written as
\[\mathbf{A} = \mathbf{L} \mathbf{L}^H\]With the positive-definiteness condition of \(l_{jj} > 0\), the decomposition is unique.
\[\begin{cases} \dfrac{1}{l_{00}} = \operatorname{rsqrt}(a_{00}), & j = 0 \\[10pt] \dfrac{1}{l_{jj}} = \operatorname{rsqrt}\biggl( a_{jj} - \sum_{k=0}^{j-1} \lvert l_{jk} \rvert^2 \biggr), & j \ge 1 \\[10pt] l_{ij} = \dfrac{1}{l_{jj}} \biggl( a_{ij} - \sum_{k=0}^{j-1} l_{ik} l_{jk}^H \biggr), & i = j + 1, \dots, n - 1 \end{cases}\]This algorithm does not explicitly calculate the diagonal elements \(l_{jj}\), thereby avoiding square root or division operations: processing each column requires only a single “reciprocal square root” operation, rather than first computing a square root followed by \(n - 1 - j\) divisions. Since the reciprocal square root can be obtained directly using the “fast inverse square root” algorithm, it is treated as a single operation rather than a combination of square root and division.
Substituting \(\mathbf{A} = \mathbf{L} \mathbf{L}^H\) into \(\mathbf{A} \hat{\mathbf{x}} = \mathbf{z}\) gives two triangular equations:
\[\begin{cases} \mathbf{L} \hat{\mathbf{z}} = \mathbf{z}\\ \mathbf{L}^H \hat{\mathbf{x}} = \hat{\mathbf{z}} \end{cases}\]Forward substitution solves the first one:
\[\begin{cases} \hat{z}_0 = \dfrac{z_0}{l_{00}}, & i = 0 \\[10pt] \hat{z}_i = \dfrac{1}{l_{ii}} \biggl( z_i - \sum_{k=0}^{i-1} l_{ik} \hat{z}_k \biggr), & i = 1, \dots, n - 1 \end{cases}\]Backward substitution solves the second one:
\[\begin{cases} \hat{x}_{n-1} = \dfrac{\hat{z}_{n-1}}{l_{n-1,n-1}}, & i = n - 1 \\[10pt] \hat{x}_i = \dfrac{1}{l_{ii}} \biggl( \hat{z}_i - \sum_{k=i+1}^{n-1} l_{ki}^H \hat{x}_k \biggr), & i = n - 2, \dots, 0 \end{cases}\]3. LDL decomposition
The same matrix can be written with a unit triangular factor and a separate diagonal:
\[\mathbf{A} = \mathbf{L} \mathbf{D} \mathbf{L}^H\]where
\[\begin{cases} l_{jj} = 1 \\[10pt] l_{ij} = 0 \quad \text{for} \quad i < j \\[10pt] \mathbf{D} = \operatorname{diag}(d_0, \dots, d_{n-1}), \qquad d_j > 0 \end{cases}\]This is a Cholesky decomposition in which the square-root terms have been factored out of the triangular factors.
\[\begin{cases} d_0 = a_{00}, & j = 0 \\[10pt] d_j = a_{jj} - \sum_{k=0}^{j-1} \lvert l_{jk} \rvert^2 d_k, & j \ge 1 \\[10pt] l_{ij} = \dfrac{1}{d_j} \biggl( a_{ij} - \sum_{k=0}^{j-1} l_{ik} l_{jk}^H d_k \biggr), & i = j + 1, \dots, n - 1 \end{cases}\]There is no square root in the algorithm. Compared with Cholesky, every inner product now also carries a real multiplication by \(d_k\), which is where the additional arithmetic goes.
With \(\mathbf{A} = \mathbf{L} \mathbf{D} \mathbf{L}^H\) the equation splits into
\[\begin{cases} \mathbf{L} \hat{\mathbf{z}} = \mathbf{z}\\ \mathbf{L}^H \hat{\mathbf{x}} = \mathbf{D}^{-1} \hat{\mathbf{z}} \end{cases}\]Forward substitution runs with a unit diagonal, so it does not need division:
\[\begin{cases} \hat{z}_0 = z_0, & i = 0 \\[10pt] \hat{z}_i = z_i - \sum_{k=0}^{i-1} l_{ik} \hat{z}_k, & i = 1, \dots, n - 1 \end{cases}\]Backward substitution folds the diagonal scaling:
\[\begin{cases} \hat{x}_{n-1} = \dfrac{\hat{z}_{n-1}}{d_{n-1}}, & i = n - 1 \\[10pt] \hat{x}_i = \dfrac{\hat{z}_i}{d_i} - \sum_{k=i+1}^{n-1} l_{ki}^H \hat{x}_k, & i = n - 2, \dots, 0 \end{cases}\]Only \(n\) reciprocals \(1 / d_i\) are required.
4. Complexity analysis
4.1 Cholesky vs LDL
| Cholesky | LDL |
|---|---|
| \(\frac{1}{l_{jj}} = \operatorname{rsqrt}( a_{jj} - \sum_{k=0}^{j-1} \lvert l_{jk} \rvert^2 )\) | \(\frac{1}{d_j} = ( a_{jj} - \sum_{k=0}^{j-1} \lvert l_{jk} \rvert^2 d_k )^{-1}\) |
| \(l_{ij} = \frac{1}{l_{jj}} ( a_{ij} - \sum_{k=0}^{j-1} l_{ik} l_{jk}^H )\) | \(l_{ij} = \frac{1}{d_j} ( a_{ij} - \sum_{k=0}^{j-1} l_{ik} l_{jk}^H d_k )\) |
| \(\hat{z}_i = \frac{1}{l_{ii}} ( z_i - \sum_{k=0}^{i-1} l_{ik} \hat{z}_k )\) | \(\hat{z}_i = z_i - \sum_{k=0}^{i-1} l_{ik} \hat{z}_k\) |
| \(\hat{x}_i = \frac{1}{l_{ii}} ( \hat{z}_i - \sum_{k=i+1}^{n-1} l_{ki}^H \hat{x}_k )\) | \(\hat{x}_i = \frac{\hat{z}_i}{d_i} - \sum_{k=i+1}^{n-1} l_{ki}^H \hat{x}_k\) |
Step by step, Cholesky differs from LDL as follows:
- The diagonal pivot operation reduces the number of real multiplications by \(\frac{n(n-1)}{2}\) and divisions by \(n\), at the cost of adding \(n\) reciprocal square root operations.
- The off-diagonal column update operation reduces the number of real-complex multiplications by \(\frac{n(n-1)(n-2)}{6}\), because the inner product terms do not contain the factor \(d_k\).
- The forward substitution operation adds \(n\) real-complex multiplications, corresponding to a diagonal scaling for each component.
- The computational cost of the backward substitution operation is the same.