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Assume we know that square matrices $A$ and $(I-AD)^{-1}$ are invertible and also $D$ is a diagonal matrix. Also assume that $A$ is a symmetric matrix. My question is when we can express $(I-AD)^{-1}$ as a function of $A, D, A^{-1}, D^{-1}, (I-A)^{-1}$ (without terms $(A-D)^{-1}$ or $(A^{-1}-D)^{-1}$) ?

For example when matrix A is rank 1, then we have:

$(I-AD)^{-1}=I+\frac{1}{1-tr(AD)} AD$. As you can see if A is rank 1 then we can do this easily.

The only related paper I found is a paper by Kenneth S. Miller, but it is not useful for higher rank matrices.

I know it might be very hard for general matrix $A$ but can it be done for special cases where for instance matrix A is positive semidefinite? Any comment is highly appreciated.

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Thanks for pointing this out, but I believe by using this theorem still term $(I-AD)^{-1}$ will show up. – user54626 Feb 14 '13 at 0:23
It's not clear to me what you mean by "a function of $A,D,A^{-1},...$ etc, because $(I - AD)^{-1}$ to me is already written as a function of those things. – Christopher A. Wong Feb 14 '13 at 0:36
For example, we can rewrite this as $(A^{-1} - D)^{-1}A^{-1}$, but that doesn't exactly simplify anything in particular. – Christopher A. Wong Feb 14 '13 at 0:37
I'm not sure what your context is, since you are mentioning $(I-A)^{-1}$ while the invertibility of this matrix does not follow from your assumption. Would you consider something along the line of $(I - AD)^{-1} = I + (AD) + (AD)^2 + \cdots$ whenever $||AD|| < 1$ as a satisfactory answer? – user27126 Feb 16 '13 at 5:40

I have an answer to the following question : Is there a four-variable (not necessarily commutative) polynomial $f$ such that the identity

$$ (I-AD)^{-1}=f((I-A)^{-1},A,D,A^{-1},D^{-1}) \tag{1} $$

holds, whenever $A$ is symmetric positive definite, and $D$ is invertible and diagonal ?

The answer is NO. Indeed, this is already impossible when $n=2$ and $$D=\left(\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}\right).$$ If we write

$$ A=\left(\begin{matrix} a & b \\ b & a \end{matrix}\right) $$

with $a\gt 0$ and $a\gt b$, then

$$ \det(A)=a^2-b^2, \det(I-A)=a^2-2a-b^2+1, \det(I-AD)=6a^2-5a-6b^2+1 $$

Thus the RHS in (1) will always have a denominator of the form

$$ (a^2-b^2)^p (a^2-2a-b^2+1)^q, $$

where $p$ and $q$ are integers. Now the LHS in (1) will always a denominator of the form $(6a^2-5a-6b^2+1)^r$. Since we have three distinct irreducible polynomials in ${\mathbb Q}[a,b]$ here, the denominators will never coincide.

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Very nice, I'm curious if it's possible to see the infinite series solution 1/(1-x) = 1 + x + ... Somehow in determinant polynomial space? – kbb Feb 19 '13 at 22:29

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