Is there a linear decomposition of the Hadamard inverse of the sum of two matrices?

Question

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 $B$.

Answer 1

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 $$ A^{\circ(-1)}=\pmatrix{1&1\\1&1}\text{ and } \quad B^{\circ(-1)}=\pmatrix{1&2\\3&4}. $$

Answer 2

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.