Suppose $A$ is a $q\times p$ matrix, $B$ is $q \times p$ matrix, $A_j$ is the jth column of $A$, and $B_j$ is the jth column of $B$. The following sum of $l_1$ norms, where "$\circ$" is the Frobenius norm
$\sum_{j=1}^p |A_j \circ B_j|$
gives the sum of the absolute values of the components of $A \circ B$, the Hadamard product of matrices $A$ and $B$. I am trying to relate this to a trace. I know that
$\sum_{i=1}^q \sum_{j=1}^p (A \circ B)_{ij} = trace(A \cdot B^\top) = trace(B \cdot A^\top)$
where $A^\top$ is the transpose of $A$. That is, the sum of all the values of the Hadamard product equals the above trace values. And if all the components of $A$ and $B$ happen to be positive, then $\sum_{j=1}^p |A_j \circ B_j| = trace(A \cdot B^\top) = trace(B \cdot A^\top)$
Any ideas?
The colon product is a convenient way to write the trace $$B:A = A:B = \operatorname{Tr}\left(A^TB\right) =\sum_i\sum_j A_{ij}B_{ij}$$ Define an element-wise sign function $$\eqalign{ \operatorname{sign}(X_{ij}) &= \begin{cases} +1\quad&{\rm if\;}X_{ij}\ge 0 \\ -1\quad&{\rm otherwise} \\ \end{cases} \\ S &= \operatorname{sign}(X) \\ }$$ and use it to write the $\ell_1$-norm as $$\eqalign{ \|X\|_1 &= S:X \\ &= \operatorname{Tr}(S^TX) \\ &= \operatorname{Tr}(X^TS) \\ }$$ If all elements of the $X$ matrix are positive, then $S$ is the all-ones matrix and $$\|X\|_1 = \operatorname{Tr}\left({\tt1}^TX\right)$$ And of course, everything can be expressed in terms of $A$, $B$, and the Hadamard product. $$\eqalign{ X &= A\circ B \\ S &= \operatorname{sign}(A)\circ\operatorname{sign}(B) \\ }$$