Approximating Hessian with BFGS for a matrix variable

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So as we know the approximation of the inverse of the hessian matrix using the BFGS method
is calculated with the following formulas :



$$q_k+1 = (I-p_k s_k (y_k)^T)q_k(I- p_k y_k (s_k)^T) + p_k s_k (s_k)^T$$
$$p_k = 1/y_k(y_k)^T$$
$$s_k = x_k+1 - x_k$$
$$y_k = nabla f(x_k+1) - nabla f(x_k)$$ Where $x$ is the input, $k = 1,2,3 ... $ the iteration number, $I$ identity matrix and $q_0 = I$ too.



Given $x_0 = 60x80$ matrix, calculating the above given BFGS equations won't work because of the matrix multiplication (with $I,q_0 = 80x80$ matrices)



what am i doing wrong ?




matrices optimization convex-optimization numerical-optimization hessian-matrix




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    write $x$ as a vector of length 4800
    – LinAlg
    Jul 23 at 12:43










  • @LinAlg $x$ is the input it differ
    – james
    Jul 23 at 13:19















up vote
0
down vote

favorite












So as we know the approximation of the inverse of the hessian matrix using the BFGS method
is calculated with the following formulas :



$$q_k+1 = (I-p_k s_k (y_k)^T)q_k(I- p_k y_k (s_k)^T) + p_k s_k (s_k)^T$$
$$p_k = 1/y_k(y_k)^T$$
$$s_k = x_k+1 - x_k$$
$$y_k = nabla f(x_k+1) - nabla f(x_k)$$ Where $x$ is the input, $k = 1,2,3 ... $ the iteration number, $I$ identity matrix and $q_0 = I$ too.



Given $x_0 = 60x80$ matrix, calculating the above given BFGS equations won't work because of the matrix multiplication (with $I,q_0 = 80x80$ matrices)



what am i doing wrong ?




matrices optimization convex-optimization numerical-optimization hessian-matrix




share|cite|improve this question

















  • 1




    write $x$ as a vector of length 4800
    – LinAlg
    Jul 23 at 12:43










  • @LinAlg $x$ is the input it differ
    – james
    Jul 23 at 13:19













up vote
0
down vote

favorite









up vote
0
down vote

favorite











So as we know the approximation of the inverse of the hessian matrix using the BFGS method
is calculated with the following formulas :



$$q_k+1 = (I-p_k s_k (y_k)^T)q_k(I- p_k y_k (s_k)^T) + p_k s_k (s_k)^T$$
$$p_k = 1/y_k(y_k)^T$$
$$s_k = x_k+1 - x_k$$
$$y_k = nabla f(x_k+1) - nabla f(x_k)$$ Where $x$ is the input, $k = 1,2,3 ... $ the iteration number, $I$ identity matrix and $q_0 = I$ too.



Given $x_0 = 60x80$ matrix, calculating the above given BFGS equations won't work because of the matrix multiplication (with $I,q_0 = 80x80$ matrices)



what am i doing wrong ?




matrices optimization convex-optimization numerical-optimization hessian-matrix




share|cite|improve this question













So as we know the approximation of the inverse of the hessian matrix using the BFGS method
is calculated with the following formulas :



$$q_k+1 = (I-p_k s_k (y_k)^T)q_k(I- p_k y_k (s_k)^T) + p_k s_k (s_k)^T$$
$$p_k = 1/y_k(y_k)^T$$
$$s_k = x_k+1 - x_k$$
$$y_k = nabla f(x_k+1) - nabla f(x_k)$$ Where $x$ is the input, $k = 1,2,3 ... $ the iteration number, $I$ identity matrix and $q_0 = I$ too.



Given $x_0 = 60x80$ matrix, calculating the above given BFGS equations won't work because of the matrix multiplication (with $I,q_0 = 80x80$ matrices)



what am i doing wrong ?





matrices optimization convex-optimization numerical-optimization hessian-matrix





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 12:46









Rodrigo de Azevedo

12.6k41751




12.6k41751









asked Jul 23 at 11:15









james

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256







  • 1




    write $x$ as a vector of length 4800
    – LinAlg
    Jul 23 at 12:43










  • @LinAlg $x$ is the input it differ
    – james
    Jul 23 at 13:19













  • 1




    write $x$ as a vector of length 4800
    – LinAlg
    Jul 23 at 12:43










  • @LinAlg $x$ is the input it differ
    – james
    Jul 23 at 13:19








1




1




write $x$ as a vector of length 4800
– LinAlg
Jul 23 at 12:43




write $x$ as a vector of length 4800
– LinAlg
Jul 23 at 12:43












@LinAlg $x$ is the input it differ
– james
Jul 23 at 13:19





@LinAlg $x$ is the input it differ
– james
Jul 23 at 13:19
















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