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[e, lambda] = ckbs_tridiag_solve_b(b, c, r)
e
:
\[
A * e = r
\]
where the symmetric block tridiagonal matrix
A
is defined by
\[
A =
\left( \begin{array}{ccccc}
b_1 & c_2^\R{T} & 0 & \cdots & 0 \\
c_2 & b_2 & c_3^\R{T} & & \vdots \\
\vdots & & \ddots & & \\
0 & \cdots & & b_N & c_N
\end{array} \right)
\]
The routine ckbs_tridiag_solve
solves the same problem.
The difference is that this routine runs the reduction starting
at the last block rather than the first block of the matrix,
which is more stable for this application.
b
is a three dimensional array,
for
k = 1 , \ldots , N
\[
b_k = b(:,:,k)
\]
and
b
has size
n \times n \times N
.
c
is a three dimensional array,
for
k = 2 , \ldots , N
\[
c_k = c(:,:,k)
\]
and
c
has size
n \times n \times N
.
r
is an
(n * N) \times m
matrix.
e
is an
(n * N) \times m
matrix.
lambda
is a scalar equal to the
logarithm of the determinant of
A
.
ckbs_tridiag_solve_b
.
It returns true if ckbs_tridiag_solve_b
passes the test
and false otherwise.