Here is an example of how the effect of columnwise rotation occurs. See the protocol using my MatMate-program. The "rot"-command performs a column-wise rotation to triangular shape, where the order of the invovled rows (the rotation-citeria) is given as list, and the triangular form is so just permuted. However, the permutation includes also a re-computation of the diagonal (and the other entries, too)
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;MatMate-Listing vom:06.12.2011 19:40:03
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[1] A = mk(4,4,6,3,4,8, 3,6,5,1, 4,5,10,7, 8,1,7, 25)
[2] L = cholesky(A)
A :
6.00 3.00 4.00 8.00
3.00 6.00 5.00 1.00
4.00 5.00 10.00 7.00
8.00 1.00 7.00 25.00
L :
2.45 0.00 0.00 0.00
1.22 2.12 0.00 0.00
1.63 1.41 2.31 0.00
3.27 -1.41 1.59 3.13
[3] M = rot(L,"drei",1´2´3´4)
Command [3] does not change anything because the order given is the default order. In
[4] M = rot(L,"drei",2´3´4´1)
M :
1.22 0.62 1.38 -1.48
2.45 0.00 0.00 0.00
2.04 2.42 -0.00 0.00
0.41 2.55 4.28 -0.00
[5] M = rot(L,"drei",3´4´1´2)
M :
1.26 1.16 1.75 0.00
1.58 -0.56 0.94 1.52
3.16 -0.00 -0.00 -0.00
2.21 4.48 0.00 -0.00
in command [6] we find the solution with the number 5 in the first column:
[6] M = rot(L,"drei",4´1´2´3)
M :
1.60 1.85 -0.00 -0.00
0.20 1.44 1.97 0.00
1.40 0.95 1.70 -2.06
5.00 0.00 -0.00 -0.00
After that I found the "modified cholesky" solution by the rotation in the following order:
[7] M = rot(L,"drei",4´3´2´1)
M :
1.60 0.62 0.92 1.48
0.20 1.66 1.79 0.00
1.40 2.84 0.00 -0.00
5.00 0.00 0.00 0.00
only that the rows are permuted.
So we can even systematically recover the original vanilla-cholesky matrix from that of the modified-cholesky-procedure: if we have r rows, then we need at most r(r-1)/2 rotations to triangular shape to recover the original vanilla-cholesky matrix.
[update 2] A remark about "rotation of columns".
Rotation of columns can be thought as postmultiplication of the cholesky
L by some rotation-matrix
T.
T itself is generated by successively rotating pairs of columns, so
T can be seen as product of elementary rotation-matrices $\small t_{k,j}=\begin{array} {rrrr} \cos(\varphi_{k,j}) & \sin(\varphi_{k,j}) \\ -\sin(\varphi_{k,j}) & \cos(\varphi_{k,j}) \end{array} $ where that
k,j'th rotation-parameters are inserted in the
ID-matrix at the appropriate pair of rows
k and
j and the same columns. Then $\small T = t_{1,2} \cdot t_{1,3} \cdot t_{1,4} \cdot \ldots t_{2,3} \cdot t_{2,4} \cdot \ldots t_{3,4} \cdot \ldots \cdot t_{n-1,n} $ where
n is the number of columns of the matrix which is to be rotated. In particular,
T is
not a permuationmatrix!
The different rotations mentioned in the example above occur, because for the triangular rotation we need to find a rotationmatrix $\small T_1 $ which rotates the matrix
L such that the entries in row
1 are collected in column
1, call this version of
L $\small L_1$. Next we need to find the rotationmatrix $\small T_2 $ which rotates then $\small L_1$ such that the entries in row
2 are collected in column
2 but where we do not touch the column
1. Save this in matrix $\small L_2$ and so on until $\small L_n = T_1 \cdot T_2 \cdot T_3 \cdot \ldots \cdot T_n $ is a triangular matrix. The different version in the example above is then simply, that the order of the $\small T_k $ in that product is changed (according to the list, given in the
rot(...)-command in MatMate. (you may try this using MatMate yourself)
[update]
The "modified cholesky" seems to proceed by extracting the row/column (thr "factor") which has the highest entry on the diagonal ("variance") first . Then proceeds with the reduced matrix after extraction of the next row/column. This explains the final permutation order between the rows 1 and 2, which have initially the same diagonal value 6, and the factor in row 1 has even a greater "individual variance" than the factor in row 2. See the following protocol, where I used a copy "chk" of the original matrix
A and proceeded to extract always that factor with the highest variance in the (residual) covariance-matrix (MatMate-command "RemVar" ("RemoveVariance of one factor"). Negative signs at the zero is spurious numerical machine-epsilon:
[24] chk = A
chk :
6.00 3.00 4.00 8.00
3.00 6.00 5.00 1.00
4.00 5.00 10.00 7.00
8.00 1.00 7.00 25.00
[25] chk = remvar(chk,4)
chk :
3.44 2.68 1.76 0.00
2.68 5.96 4.72 0.00
1.76 4.72 8.04 0.00
0.00 0.00 0.00 0.00
[26] chk = remvar(chk,3)
chk :
3.05 1.65 -0.00 0.00
1.65 3.19 -0.00 0.00
-0.00 -0.00 -0.00 0.00
0.00 0.00 0.00 0.00
[27] chk = remvar(chk,2)
chk :
2.20 0.00 0.00 0.00
0.00 0.00 0.00 0.00
0.00 0.00 -0.00 0.00
0.00 0.00 0.00 0.00
The $LDL^T$ decomposition of an SPSD matrix cannot be unique. If
$$
A=LDL^T
$$
with
$$\tag{1}
D=\begin{bmatrix}D_{11}&0\\0&0\end{bmatrix}, \quad
L=\begin{bmatrix}L_{11}&0\\L_{21}&L_{22}\end{bmatrix},
$$
where $D_{11}$ is nonsingular (with positive diagonal entries), then it is easy to see that the sub-matrix $L_{22}$ can be in fact chosen arbitrarily. Generally, you would need to consider a pivoted factorisation leading to
$$\tag{2}
\Pi^TA\Pi=LDL^T
$$
with $L$ and $D$ of the form (1) and some permutation matrix $\Pi$, because accepting a zero pivot would make the remainder of the factorisation algorithm undefined.
Assume that you have a factorisation (1) obtained (by luck) without pivoting (or consider $\Pi^TA\Pi$ instead of $A$) and define $A^+=L^{-T}D^+L^{-1}$. It is easy to verify that
$$
A^+=\begin{bmatrix}L_{11}^{-T}D_{11}^{-1}L_{11}^{-1}&0\\0&0\end{bmatrix},
$$
so $A^+$ is unique as it does not depend on the "non-unique block" $L_{22}$. So in fact
$$
A^{+}=\begin{bmatrix}A_{11}^{-1}&0\\0&0\end{bmatrix},
$$
where $A_{11}$ is the leading principal sub-matrix of $A$ (of the dimension equal to the rank of $A$ consistent with the partitioning of the factors in (1)).
You might want to note that $A^{+}$ defined this way is not the Moore-Penrose pseudo inverse, since it generally $AA^{+}$ and $A^+A$ are not symmetric. On the other hand, the matrix $A^{+}$ as you defined it would form a so-called generalised reflexive inverse, or a (1,2)-generalised inverse (since it satisfies the first two of the four conditions defining the unique Moore-Pseudo inverse).
If you insists to compute the Moore-Penrose pseudo-inverse from the $LDL^T$ factorisation, consider (as before with luck or pivoting) that you have (1) and write $A$ as
$$
A=\tilde{L}D_{11}\tilde{L}^{T},
$$
where $\tilde{L}^T=[L_{11}^T,L_{21}^T]$.
Since $D_{11}$ and $\tilde{L}$ have full rank we can write
$$
A^{\dagger}=(\tilde{L}D_{11}\tilde{L}^T)^{\dagger}=(\tilde{L}^{\dagger})^TD_{11}^{-1}\tilde{L}^{\dagger},
$$
where
$$
\tilde{L}^{\dagger}=(\tilde{L}^T\tilde{L})^{-1}\tilde{L}^T
=(L_{11}^TL_{11}+L_{21}^TL_{21})^{-1}[L_{11}^T,L_{21}^T].
$$
Hence we obtain quite an awful expression
$$
A^{\dagger}=\tilde{L}\tilde{D}_{11}^{-1}\tilde{L}^{T},
\quad
\tilde{D}_{11}=\tilde{L}^T\tilde{L}D_{11}\tilde{L}^T\tilde{L}=(L_{11}^TL_{11}+L_{21}^TL_{21})D_{11}(L_{11}^TL_{11}+L_{21}^TL_{21}).
$$
Best Answer
Yes, there is a direct equivalence. Let $K$ denote the matrix of $1$'s on the "perdiagonal", i.e. $$ K = \pmatrix{&&&1\\&&1\\&\cdots\\1}, $$ with the blank entries equal to zero. Note that $K = K^T = K^{-1}$, and that $M$ is upper triangular if and only if $KMK$ is lower triangular. In general, the change from $M$ to $KMK$ consists of reversing the order of both the rows and the columns of $M$.
Suppose that we want a decomposition of the form $\Omega = T^T D^{-1}T$. Begin by computing a decomposition $K\Omega K = L^TD_0 L$. We have $$ \begin{align} K\Omega K &= L^TD_0 L \implies\\ \Omega &= KL^TD_0 LK \\ & = KL^T(KK)D_0(KK)LK \\ & = (KL^TK)(KD_0K)(KLK) \\&= (KLK)^T (KD_0K)(KLK). \end{align} $$ Note that because $D_0$ is diagonal, $KD_0K$ is also diagonal (with its diagonal reverse relative to those of $D_0$). Taking $T = KLK$ and $D = (KD_0K)^{-1} = KD_0^{-1}K$, we have $\Omega = T^T D^{-1}T$, which is what we wanted.