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Give a diagonal perturbation matrix $D$ (which is not an identity matrix), is there a simple way to compute

$$\text{trace}((A+D)^{-1}A)$$

Or is there a good approximation?

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1 Answer 1

If $A$ is invertible and $\|D\|$ is small, $(A+D)^{-1}A=(I+A^{-1}D)^{-1}\approx I-A^{-1}D$ and hence $\operatorname{trace}\left((A+D)^{-1}A\right)\approx n-\operatorname{trace}(A^{-1}D)$.

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I was just about to say the same with one small addition. For the last term, I think, you can derive some estimate $|\operatorname{Tr}(A^{-1}D)|=|A^{-T}:D|\leq\|A^{-1}\|_F\|D\|_F$, where $:$ is the Frobenius inner product and $\|\cdot\|_F$ is the Frobenius norm. –  Elmar Zander Mar 9 '13 at 9:10
    
@ElmarZander Thanks, it's a useful addition. –  user1551 Mar 9 '13 at 9:27
    
@user1551, I am just curious. What does it mean when you put $\approx$ between two matrices? –  Easy Mar 9 '13 at 11:14
    
@user60079 It means the matrix norm of their difference is small. –  user1551 Mar 9 '13 at 11:37

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