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Given are two matrices:

$\bf A, \bf B$

We know that matrices $\bf A \neq \bf B$ are invertable, symmetric, positive-definite and of full rank. Is it possible to give the formula for following sum of these matrices:

$[\bf A + \lambda\bf B]^{-1} = ?$

where $\lambda$ is a scalar such as $0 < \lambda < 1$.

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  • $\begingroup$ It is possible for $A$ and $B$ to be invertible without $A+\lambda B$ being so. For instance take $A=B$ and $\lambda =-1$. $\endgroup$
    – Git Gud
    Mar 2, 2013 at 20:23
  • $\begingroup$ OK, I've edited the question $\endgroup$
    – Gacek
    Mar 2, 2013 at 20:26
  • $\begingroup$ Still the same problem, you could have $B=-\frac{1}{\lambda}A$. $\endgroup$
    – Integral
    Mar 2, 2013 at 20:28
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    $\begingroup$ Have you seen the Binomial Inverse Theorem? $\endgroup$
    – Josephine Moeller
    Mar 2, 2013 at 20:28
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    $\begingroup$ The space of positive definite matrices is convex. So $A + \lambda B$ is invertible for $\lambda > 0$. $\endgroup$
    – achille hui
    Mar 2, 2013 at 20:40

2 Answers 2

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Assuming that $A+\lambda B$ is also invertible, you can use the Binomial Inverse Theorem:

$[A+\lambda B]^{-1} = A^{-1}-\lambda A^{-1}(I + \lambda BA^{-1})^{-1}BA^{-1}$

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Use Woodbury matrix identity:

$$ \left(A+UCV \right)^{-1} = A^{-1} - A^{-1}U \left(C^{-1}+VA^{-1}U \right)^{-1} VA^{-1}. $$

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