There is a objective function:

$f(W)$ = $||W||_{2,1}$

For any element $w_{ab}$ in $W$, we apply $F_{w_{ab}}$ to denote the part of $f(W)$ which is only related to $w_{ab}$.


$F''_{w_{ab}}=(D-D^{3}(W\odot W))_{aa}$, (the main problem)

where $D_{ii}$=$\frac{1}{||w^i||_2}$, $\odot$ denotes the element-wise multiplication.

About $F_{w_{ab}}$: why the second order derivative of F is equation (28)?

Note that: The norm $\|\cdot\|_{2,1}$ of a matrix $W\in\mathbb{R}^{n\times m}$ is defined as

$$ \Vert W \Vert_{2,1} = \sum_{i=1}^n \Vert w^{i} \Vert_2 = \sum_{i=1}^n \left( \sum_{j=1}^m |w_{ij}|^2 \right)^{1/2} $$ where $w^i$ denotes $i^{th}$ row of $W$, $w_{ij}$ denotes a element of $W$.

How to solve $F''_{w_{ab}}$? I want to know the detailed calculation process of solving the above formula.

There some more explicit definition in the following papers (I gave the exact location.)

Some Related Papers:

Efficient and Robust Feature Selection via Joint $l_{2,1}$-Norms Minimization

Graph Regularized Nonnegative Matrix Factorization for Data Representation (Look Page 10, APPENDIX A, PROOFS OF THEOREM 1, formulas (26), (27) and (28))

In Nonnegative matrix factorization by joint locality-constrained and $l_{2,1}$-norm regularization, (Look Page 7, the Proof of Lemma 1: Here, how to obtain the result of $F^{''}_{v_{ab}}$?)

Thank you all for your help.

  • 1
    $\begingroup$ Can you define $F_{w_{ab}}$ more explicitly? $\endgroup$ – littleO May 16 at 7:46
  • $\begingroup$ I think the chance of getting a helpful response would be higher if the definition of $F_{w_{ab}}$ were included in the question. $\endgroup$ – littleO May 16 at 8:03
  • $\begingroup$ This detailed answer shows how to calculate the derivative of the mixed norm. The only difference is that you use the symbol $\|W\|_{2,1}$ whereas he uses $\|W\|_{1,2}$. Oddly, neither symbol corresponds with the standard usage, at least according to Wikipedia. $\endgroup$ – greg May 17 at 15:45
  • $\begingroup$ Since $D\in{\mathbb R}^{n\times n}$ and $W\in{\mathbb R}^{n\times m},\,$ the product $D^3(W\odot W)\in{\mathbb R}^{n\times m}\,$ is well defined. However, your "main" equation $$F_{ab}'' = D-D^3(W\odot W)$$ is dimensionally incompatible. You cannot subtract a rectangular matrix from a square one. It only exists if $m=n$. $\endgroup$ – greg May 20 at 20:22

Define $$\eqalign{ b &= (W\odot W)\,1 \implies db=2(W\odot dW)1 \cr B &= {\rm Diag}(b),\quad D = B^{-1/2},\quad E=e_a\varepsilon_b^T \cr }$$ Write the $\{2,1\}$ norm in terms of $b$ and find its gradient. $$\eqalign{ f &= \|W\|_{2,1} \cr&= b^{\odot 1/2}:1 \cr df &= \tfrac{1}{2}b^{\odot-1/2}:db \cr &= b^{\odot-1/2}:(W\odot dW)1 \cr &= b^{\odot-1/2}1^T:(W\odot dW) \cr &= (b^{\odot-1/2}1^T\odot W):dW \cr &= B^{-1/2} W:dW \cr F'=\frac{\partial f}{\partial W} &= B^{-1/2} W = DW \cr F_{ab}' &= e_a^TF'\varepsilon_b = E:DW \cr }$$ Now find the gradient of this last quantity. $$\eqalign{ dF_{ab}' &= E:(D\,dW+dD\,W) \cr &= DE:dW + EW^T:dB^{-1/2} \cr &= DE:dW - \tfrac{1}{2}EW^T:B^{-3/2}dB \cr &= DE:dW - \tfrac{1}{2}{\rm diag}(D^3EW^T):db \cr &= DE:dW - {\rm diag}(D^3EW^T):(W\odot dW)1 \cr &= DE:dW - \big({\rm diag}(D^3EW^T)1^T\big)\odot W:dW \cr &= DE:dW - {\rm Diag}(D^3EW^T)W:dW \cr &= \Big(DE - D^3\,{\rm Diag}(EW^T)W\Big):dW \cr F_{ab}'' = \frac{\partial F_{ab}'}{\partial W} &= DE - D^3\,{\rm Diag}(EW^T)W \cr }$$ In the above, $1$ is a vector with all elements equal to one, $e_a$ is a vector from the standard basis for ${\mathbb R}^{n}$, the basis vector $\varepsilon_b$ is from ${\mathbb R}^{m}$, a colon denotes the trace/Frobenius product, e.g. $$A:B={\rm Tr}(A^TB)$$ the symbol $\odot$ denotes the elementwise/Hadamard product, and Hadamard powers are denote as $$A^{\odot 3}=A\odot A\odot A$$ The function $a={\rm diag}(A)$ extracts the diagonal elements of $A$ into a vector $a$
while the function ${\rm Diag}(A)$ sets the off-diagonal elements of $A$ to zero.

  • $\begingroup$ Thank you very much. $\endgroup$ – learner May 21 at 1:17
  • $\begingroup$ $b^{\odot -1/2}1^T \odot W = B^{-1/2}W$? I think that this is wrong by computing. $\endgroup$ – learner May 21 at 12:35
  • $\begingroup$ Given any two vectors $(a,c)$ and a matrix $X$ convince yourself that $$A = {\rm Diag}(a) \\C = {\rm Diag}(c) \\AXC = ac^T\odot X$$ Then let $\,\,c=1,\,\,a=b^{\odot p},\,\,X=W$ $\endgroup$ – greg May 21 at 13:47
  • $\begingroup$ OK,I miscalculated. You're right. $\endgroup$ – learner May 22 at 1:22
  • $\begingroup$ I have three problems:First, $df=\tfrac{1}{2}b^{\odot-1/2}:db$? Second, $\tfrac{1}{2}EW^T:B^{-3/2}dB=\tfrac{1}{2}{\rm diag}(D^3EW^T):db$? Third, how to transform $E$ from Diag() to outside in $D^3\,{\rm Diag}(EW^T)W$? $\endgroup$ – learner May 22 at 6:19

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