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Let the matrix $$\Gamma = \alpha A + (1-\alpha)B$$ where $B$ is a square symmetric matrix, $A = c\ ee'$, where $e$ is a vector of ones, and $c$ a positive constant and $0 < \alpha < 1$.

The Hadamard inverse is defined as $$[\Gamma^{\circ(-1)}]_{ij} = 1/[\Gamma]_{ij} $$ (e.g. Reams).

Is there a linear decomposition of $\Gamma^{\circ(-1)}$ in terms of $A$ and $B$, assuming that $\Gamma^{\circ(-1)}$ is well defined?

Ideally, I am looking for a solution of the form $$\Gamma^{\circ(-1)}= a A^{\circ(-1)} + bB^{\circ(-1)}$$.

with $a$ and $b$ being some function of $\alpha$, $A$ and $B$.

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    $\begingroup$ The Hadamard inverse fails to be defined for a certain set of values of $\alpha$ that depend on the individual entries of $B$. I think this shows that an expression of the kind you are asking for cannot exist. (It may depend on just how wild (and therefore useless) you are willing to allow your "linear decomposition" to be; the question is not very clear about the form you are looking for, but it seems that for any reasonable form the answer must be "no".) $\endgroup$ Commented Aug 21, 2015 at 9:10
  • $\begingroup$ Thanks for the answer @Marc van Leeuwen but I do not understand why you think it is not possible. In the problem I am looking at $0 < \alpha < 1$. In the best of worlds, I am looking for a solution of the form $$a A^{\circ(-1)} + bB^{\circ(-1)}$$. $\endgroup$
    – Daniel
    Commented Aug 21, 2015 at 9:30

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Do a simple example, and you will see that a linear combination of $A^{\circ(-1)}$ and $B^{\circ(-1)}$ does not suffice. The first one that comes to my mind: $\alpha=\frac12$ and $$ A=\pmatrix{1&1\\1&1}, \quad B=\pmatrix{1&1/2\\1/3&1/4}; $$ now $$ \Gamma=\pmatrix{1&3/4\\2/3&5/8}=\frac1{24}\pmatrix{24&18\\16&15} $$ and as you can easily verify this is not a linear combination of $$ A^{\circ(-1)}=\pmatrix{1&1\\1&1}\text{ and } \quad B^{\circ(-1)}=\pmatrix{1&2\\3&4}. $$

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  • $\begingroup$ thank you @math.stackexchange.com/users/18880/marc-van-leeuwen. I forgot to mentioned that, in my case $B$ is symmetric. Just edited the question to add that. $\endgroup$
    – Daniel
    Commented Aug 21, 2015 at 20:40
  • $\begingroup$ Even if $B$ was symmetric in a similar example I can also see it is not possible. Say $$B=\frac{1}{24}\pmatrix{24&18\\18&15}=\frac{3}{24}\pmatrix{8&6\\6&5}$$ and so $$a\pmatrix{1&1\\1&1}+b\pmatrix{1&2\\2&4} \neq \pmatrix{8&6\\6&5}$$ for any $a, b$. Thanks a lot! now I can stop looking for it! :) $\endgroup$
    – Daniel
    Commented Aug 21, 2015 at 21:00
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You're looking for an identity $$ \dfrac{1}{\alpha c + (1-\alpha) x} = \dfrac{a}{c} + \dfrac{b}{x}$$ that should hold where $c$ is an entry of $A$ and $x$ an entry of $B$. Clearing out denominators and collecting terms in $x$, this can be written as $$ \left( a\alpha-a \right) {x}^{2}+ \left( -a\alpha\,c+\alpha\,bc-bc+c \right) x-\alpha\,b{c}^{2} = 0$$ The only way this can be true for more than two $x$ is if all coefficients are $0$. It's easy to check that this can't work if $c \ne 0$ and $\alpha \ne 0, 1$. So any matrix $B$ with more than two distinct entries is a counterexample.

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