## 4 Recursive Solution of a Linear Equation with a Symmetric One-Block-Banded Matrix  
Before embarking on the actual discussion of the weighted least-squares approach to state estimation, an algorithm for the recursive solution of a linear equation needs to be briefly reviewed. Let the linear equation be given by  
\[\mathcal{V}=\mathcal{M}\mathcal{X}, \tag{43}\]  
where the square matrix, \(\mathcal{M}\), has a particular structure, known as symmetric one-block-banded matrix structure. This structure is given by  
\[\mathcal{M}= \begin{pmatrix}\mathsf{M}_{1}&\mathsf{W}_{2}&0&\dots&0&0\\ \mathsf{W}_{2}^{\mathsf{T}}&\mathsf{M}_{2}&\mathsf{W}_{3}&\dots&0&0\\ 0&\mathsf{W}_{3}^{\mathsf{T}}&\mathsf{M}_{3}&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&\mathsf{M}_{N-1}&\mathsf{W}_{N}\\ 0&0&0&\dots&\mathsf{W}_{N}^{\mathsf{T}}&\mathsf{M}_{N}\end{pmatrix} , \tag{44}\]  
where \(\mathsf{M}_{t}\), \(t=1,\dots,N\), denotes a set of invertible symmetric positive definite \((m\times m)\)-dimensional square matrices, and \(\mathsf{W}_{t}\), \(t=2,\dots,N\), denotes another set of \((m\times m)\)-dimensional square matrices. For the vectors \(\mathcal{V}\) and \(\mathcal{X}\), a similar partitioning as for \(\mathcal{M}\) shall be defined:  
\[\mathcal{V}= \begin{pmatrix}\mathbf{v}_{1}\\ \vdots\\ \mathbf{v}_{N}\end{pmatrix} \qquad\qquad\mathcal{X}= \begin{pmatrix}\mathbf{x}_ {1}\\ \vdots\\ \mathbf{x}_{N}\end{pmatrix} , \tag{45}\]  
where the dimensions of the subvectors \(\mathbf{v}_{t}\) and \(\mathbf{x}_{t}\), \(t=1,\dots,N\), shall be \(m\). The solution to Equation (43) can be based on the blockwise LDL decomposition of \(\mathcal{M}\) according to  
\[\mathcal{M}=\mathcal{LDL}\mathcal{L}^{\mathsf{T}}, \tag{46}\]  
where  
\[\mathcal{L}= \begin{pmatrix}\mathsf{I}^{(m)}&0&0&\dots&0&0\\ \mathsf{W}_{2}^{\mathsf{T}}\mathsf{Z}_{1}^{-1}&\mathsf{I}^{(m)}&0&\dots&0&0\\ 0&\mathsf{W}_{3}^{\mathsf{T}}\mathsf{Z}_{2}^{-1}&\mathsf{I}^{(m)}&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&\mathsf{I}^{(m)}&0\\ 0&0&0&\dots&\mathsf{W}_{N}^{\mathsf{T}}\mathsf{Z}_{N-1}^{-1}&\mathsf{I}^{(m)} \end{pmatrix}  \tag{47}\]  
and  
\[\mathcal{D}= \begin{pmatrix}\mathsf{Z}_{1}&0&0&\dots&0&0\\ 0&\mathsf{Z}_{2}&0&\dots&0&0\\ 0&0&\mathsf{Z}_{3}&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&\mathsf{Z}_{N-1}&0\\ 0&0&0&\dots&0&\mathsf{Z}_{N}\end{pmatrix}  \tag{48}\]  
Then, in the first step, the equation  
\[\mathcal{V}=\mathcal{M}\mathcal{X}=\mathcal{LDL}\mathcal{L}^{\mathsf{T}}\mathcal{X}\]  
is recursively solved for \(\mathcal{Z}\), which is defined by  
\[\mathcal{Z}=\mathcal{LDL}^{\mathsf{T}}\mathcal{X}, \tag{49}\]while in the second step, Equation (49) is recursively solved for \(\mathcal{X}\). The corresponding recursions were given by [PERSON] [14], as follows:  
\[1. \quad Z_{1}=M_{1} \tag{50}\] \[\quad\mathbf{z}_{1}=\mathbf{v}_{1}\] (51) \[2. \quad\text{``forward'' recursion over $t$ from $t=2$ until $t=N$:}\] \[\quad Z_{t}=M_{t}-W_{t}^{\mathsf{T}}Z_{t-1}^{-1}W_{t}\] (52) \[\quad\mathbf{z}_{t}=\mathbf{v}_{t}-W_{t}^{\mathsf{T}}Z_{t-1}^{-1} \mathbf{z}_{t-1}\] (53) \[3. \quad\mathbf{x}_{N}=Z_{N}^{-1}\mathbf{z}_{N}\] (54) \[4. \quad\text{``backward'' recursion over $t$ from $t=N-1$ until $t=1$:}\] \[\quad\mathbf{x}_{t}=Z_{t}^{-1}(\mathbf{z}_{t}-W_{t+1}\mathbf{x}_{t+1}) \tag{55}\]  
It is not hard to see that the \"forward\" recursion leads to a Kalman filter, while the \"backward\" recursion leads to a smoother. In addition to the equations given by [PERSON], another equation needs to be added to the \"backward\" recursion. For this purpose, the task of computing the inverse of the matrix \(\mathcal{M}\) explicitly is studied, which can be carried out by inverting the matrices \(\mathcal{L}\) and \(\mathcal{D}\) separately. \(\mathcal{D}\) has a block-diagonal structure, with blocks \([\mathcal{D}]_{tt}=Z_{t}\), where the index \(t\) refers to blocks. Its inverse also has a block-diagonal structure, with blocks \([\mathcal{D}^{-1}]_{tt}=Z_{t}^{-1}\).