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I have some problems in simplification of formula with Hadamard product. First, we have a formulation as $$L(D)=\mathbf{1}^T_m(x\circ(Dw)\circ(Dw)),$$ where $L(D)\in R$, $x\in R^m$, $D\in R^{m\times k}$, $w\in R^k$, $\mathbf{1}_m$ is $m$-dimension column vector with all element being $1$ and $\circ$ denotes the Hadamard product.

Since the formulation of $L(D)$ is complicated related to variable $D$, we would like to simplify it. So what is the best simplification for $L(D)$?

Furthermore, if we consider time-varying coefficients like $$\hat{L}(D) = \sum^T_{t=1} \mathbf{1}^T_m((x_t)\circ(Dw_t)\circ(Dw_t)).$$

Can we have the format of $\hat{L}(D) = trace(D A_t)$ or $\hat{L}(D) = trace(D^TD A_t)$, where $A_t$ is a function of $x_t$ and $w_t$?

I am getting stuck in simplifying $L(D)$ and $\hat{L}(D)$. If you could give me some hints on this simplification, I will be highly appreciated. Thanks in advance.

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1 Answer 1

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Not sure if this is what you are looking for but

$$ L(D) = \mathbf{1}_m^T(x\circ (Dw)\circ (Dw)) = \sum_{i=1}^m x_i [(Dw)_i]^2 = w^TD^T\text{diag}(x)Dw $$

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  • $\begingroup$ Thanks for your answer. Now the problem is $diag(x)$ in $\hat{L}(D)$. Because there is cumulative information of $x_t$ in $\hat{L}(D)$. Can we have further simplification to obtain the elegant expression of $D^TD$? Thanks again. $\endgroup$
    – aaronyxt
    Commented Dec 29, 2015 at 6:14

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