Calculation of matrix according to Schubert's method
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Schubert's method is the improvement of Broyden's method for the calculation of the quasi-newton update of a jacobian matrix, when the matrix itself is sparse. According to his paper, the update algorithm is
$$B_{n+1}=B_n - sum_{i=1}^mu_iu_i^T(B_np-y/t)frac{p_i^T}{p_i^Tp_i}$$
with $B$ a (sparse) matrix, $p$ and $y$ vectors, and $t$ a scalar. $p_i$ is defined as $p$ with the values corresponding to zero-entries in column $i$ in matrix $B$ set to zero. $u_i$ is the $i$-th column of the unit matrix of order $m$.
Based on my understanding that gives me
$$B_{n+1}=B_n - sum_{i=1}^munderbrace{underbrace{u_iu_i^T}_{=1}underbrace{(B_np-y/t)}_{text{column vector}}underbrace{frac{p_i^T}{p_i^Tp_i}}_{text{row vector}}}_{text{scalar}}$$
a scalar value, which then is added to the original matrix. That can not be correct. Thus, where did I make my mistake here?
matrices newton-raphson
asked Dec 4 at 10:50
arc_lupus
232218
add a comment |
up vote
0
down vote
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Schubert's method is the improvement of Broyden's method for the calculation of the quasi-newton update of a jacobian matrix, when the matrix itself is sparse. According to his paper, the update algorithm is
$$B_{n+1}=B_n - sum_{i=1}^mu_iu_i^T(B_np-y/t)frac{p_i^T}{p_i^Tp_i}$$
with $B$ a (sparse) matrix, $p$ and $y$ vectors, and $t$ a scalar. $p_i$ is defined as $p$ with the values corresponding to zero-entries in column $i$ in matrix $B$ set to zero. $u_i$ is the $i$-th column of the unit matrix of order $m$.
Based on my understanding that gives me
$$B_{n+1}=B_n - sum_{i=1}^munderbrace{underbrace{u_iu_i^T}_{=1}underbrace{(B_np-y/t)}_{text{column vector}}underbrace{frac{p_i^T}{p_i^Tp_i}}_{text{row vector}}}_{text{scalar}}$$
a scalar value, which then is added to the original matrix. That can not be correct. Thus, where did I make my mistake here?
matrices newton-raphson
asked Dec 4 at 10:50
arc_lupus
232218
$u_i$ is a column vector, therefore $(u_iu_i^T)$ is a matrix; definitely not equal to the scalar value $1$.
– greg
Dec 4 at 20:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Schubert's method is the improvement of Broyden's method for the calculation of the quasi-newton update of a jacobian matrix, when the matrix itself is sparse. According to his paper, the update algorithm is
$$B_{n+1}=B_n - sum_{i=1}^mu_iu_i^T(B_np-y/t)frac{p_i^T}{p_i^Tp_i}$$
with $B$ a (sparse) matrix, $p$ and $y$ vectors, and $t$ a scalar. $p_i$ is defined as $p$ with the values corresponding to zero-entries in column $i$ in matrix $B$ set to zero. $u_i$ is the $i$-th column of the unit matrix of order $m$.
Based on my understanding that gives me
$$B_{n+1}=B_n - sum_{i=1}^munderbrace{underbrace{u_iu_i^T}_{=1}underbrace{(B_np-y/t)}_{text{column vector}}underbrace{frac{p_i^T}{p_i^Tp_i}}_{text{row vector}}}_{text{scalar}}$$
a scalar value, which then is added to the original matrix. That can not be correct. Thus, where did I make my mistake here?
matrices newton-raphson
asked Dec 4 at 10:50
arc_lupus
232218
Schubert's method is the improvement of Broyden's method for the calculation of the quasi-newton update of a jacobian matrix, when the matrix itself is sparse. According to his paper, the update algorithm is
$$B_{n+1}=B_n - sum_{i=1}^mu_iu_i^T(B_np-y/t)frac{p_i^T}{p_i^Tp_i}$$
with $B$ a (sparse) matrix, $p$ and $y$ vectors, and $t$ a scalar. $p_i$ is defined as $p$ with the values corresponding to zero-entries in column $i$ in matrix $B$ set to zero. $u_i$ is the $i$-th column of the unit matrix of order $m$.
Based on my understanding that gives me
$$B_{n+1}=B_n - sum_{i=1}^munderbrace{underbrace{u_iu_i^T}_{=1}underbrace{(B_np-y/t)}_{text{column vector}}underbrace{frac{p_i^T}{p_i^Tp_i}}_{text{row vector}}}_{text{scalar}}$$
a scalar value, which then is added to the original matrix. That can not be correct. Thus, where did I make my mistake here?
matrices newton-raphson
matrices newton-raphson
asked Dec 4 at 10:50
arc_lupus
232218
asked Dec 4 at 10:50
arc_lupus
232218
asked Dec 4 at 10:50
arc_lupus
232218
asked Dec 4 at 10:50
arc_lupus
232218
asked Dec 4 at 10:50
arc_lupus
232218
232218
$u_i$ is a column vector, therefore $(u_iu_i^T)$ is a matrix; definitely not equal to the scalar value $1$.
– greg
Dec 4 at 20:47
add a comment |
$u_i$ is a column vector, therefore $(u_iu_i^T)$ is a matrix; definitely not equal to the scalar value $1$.
– greg
Dec 4 at 20:47
$u_i$ is a column vector, therefore $(u_iu_i^T)$ is a matrix; definitely not equal to the scalar value $1$.
– greg
Dec 4 at 20:47
$u_i$ is a column vector, therefore $(u_iu_i^T)$ is a matrix; definitely not equal to the scalar value $1$.
– greg
Dec 4 at 20:47
add a comment |
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$u_i$ is a column vector, therefore $(u_iu_i^T)$ is a matrix; definitely not equal to the scalar value $1$.
– greg
Dec 4 at 20:47