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I'm trying to relate the Frobenius Norm of a Hadamard Product to a trace that does not include another Hadamard Product, if possible. In other words, if A and B are (sxr) matrices, with not all positive values,

$\left||A\circ B \right||^2_F = $?

I'm trying to relate the sum of the squares of all the entries of the Hadamard product. I know that the sum of all the entries of the Hadamard product are

$\sum_i \sum_j (A\circ B)_{ij} = tr(A B^T) $

and also

$\left||A\circ B \right||^2_F = tr((A\circ B)^T(A\circ B)) $

But I am trying to get $\left||A\circ B \right||^2_F $ in the form of the trace of some combination of A and B, without a Hadamard product. Even an inequality would help. I've tried various identities and inequalities related to Hadamard Product and Frobenius Norm, but I am not having any luck.

Any suggestions would be appreciated.

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Let $$\eqalign{ D_a &= {\rm Diag}({\rm vec}(A)) \cr D_b &= {\rm Diag}({\rm vec}(B)) \cr }$$ Then $$\eqalign{ \|A\circ B\|_F^2 &= \|D_aD_b\|_F^2 = {\rm tr}(D_a^2D_b^2) \cr }$$

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  • $\begingroup$ Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working. $\endgroup$ – Rich Jul 18 '18 at 0:36
  • $\begingroup$ Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this. $\endgroup$ – greg Jul 18 '18 at 2:07
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Since $\Vert A\Vert_F^2=\text{tr}(A^TA)=\sum_{i,j}A_{ij}^2$ for any matrix $A$, we have $$\Vert A\circ B\Vert_F^2=\sum_{i,j}(A\circ B)_{ij}^2=\sum_{i,j}A_{ij}^2B_{ij}^2.$$ You can use this to get a silly lower bound on $\Vert A\circ B\Vert_F^2$ in terms of traces of $A$ and $B$ via Cauchy-Schwarz. If $A,B\in\mathbb{R}^{m\times n}$, then $\Vert A\circ B\Vert_F^2\geq {1\over mn}\text{tr}(A^TB)^2$.

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