How to get a simpler matrix representation of weighted nuclear norm?
$begingroup$
Given a matrix $Ain R^{ntimes n}$, the nuclear norm is $|A|_*=sum_{i=1}^nsigma_i(A)$, where $sigma_i(A)$ is the $i$-th singular value of $A$. Calculating $sigma_i(A)$ is very complex. But luckily, $|A|_*$ has a simpler matrix representation as trace$(sqrt{A^TA})$, which is easy to be computed.
For the weighted nuclear norm, it is defined as $sum_{i=1}^nw_isigma_i(A)$, where $w_i$ is non-negative. My question is how to get its simpler matrix representation? Here, you can also assume that $Ain R^{ntimes n}$ is a symmetric matrix.
calculus linear-algebra matrices
asked Dec 4 '18 at 2:30
oliviaolivia
791616
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add a comment |
$begingroup$
Given a matrix $Ain R^{ntimes n}$, the nuclear norm is $|A|_*=sum_{i=1}^nsigma_i(A)$, where $sigma_i(A)$ is the $i$-th singular value of $A$. Calculating $sigma_i(A)$ is very complex. But luckily, $|A|_*$ has a simpler matrix representation as trace$(sqrt{A^TA})$, which is easy to be computed.
For the weighted nuclear norm, it is defined as $sum_{i=1}^nw_isigma_i(A)$, where $w_i$ is non-negative. My question is how to get its simpler matrix representation? Here, you can also assume that $Ain R^{ntimes n}$ is a symmetric matrix.
calculus linear-algebra matrices
asked Dec 4 '18 at 2:30
oliviaolivia
791616
$endgroup$
add a comment |
$begingroup$
Given a matrix $Ain R^{ntimes n}$, the nuclear norm is $|A|_*=sum_{i=1}^nsigma_i(A)$, where $sigma_i(A)$ is the $i$-th singular value of $A$. Calculating $sigma_i(A)$ is very complex. But luckily, $|A|_*$ has a simpler matrix representation as trace$(sqrt{A^TA})$, which is easy to be computed.
For the weighted nuclear norm, it is defined as $sum_{i=1}^nw_isigma_i(A)$, where $w_i$ is non-negative. My question is how to get its simpler matrix representation? Here, you can also assume that $Ain R^{ntimes n}$ is a symmetric matrix.
calculus linear-algebra matrices
asked Dec 4 '18 at 2:30
oliviaolivia
791616
$endgroup$
Given a matrix $Ain R^{ntimes n}$, the nuclear norm is $|A|_*=sum_{i=1}^nsigma_i(A)$, where $sigma_i(A)$ is the $i$-th singular value of $A$. Calculating $sigma_i(A)$ is very complex. But luckily, $|A|_*$ has a simpler matrix representation as trace$(sqrt{A^TA})$, which is easy to be computed.
For the weighted nuclear norm, it is defined as $sum_{i=1}^nw_isigma_i(A)$, where $w_i$ is non-negative. My question is how to get its simpler matrix representation? Here, you can also assume that $Ain R^{ntimes n}$ is a symmetric matrix.
calculus linear-algebra matrices
calculus linear-algebra matrices
asked Dec 4 '18 at 2:30
oliviaolivia
791616
asked Dec 4 '18 at 2:30
oliviaolivia
791616
edited Dec 6 '18 at 2:28
olivia
asked Dec 4 '18 at 2:30
oliviaolivia
791616
asked Dec 4 '18 at 2:30
oliviaolivia
791616
asked Dec 4 '18 at 2:30
oliviaolivia
791616
791616
add a comment |
add a comment |
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