# Comparing diagonal elements between two inverse matrices

I would like to compare diagonal elements between inverse matrices. Suppose that we have three real block matrices as follows:

$$\underbrace{\begin{bmatrix}\mathbf A & \mathbf B^T \\ \mathbf B & \mathbf C \end{bmatrix}}_{\mathbf M} = \underbrace{\begin{bmatrix}\mathbf A_1 & \mathbf 0 \\ \mathbf 0 & \mathbf C_1 \end{bmatrix}}_{\mathbf M_1} + \underbrace{\begin{bmatrix}\mathbf A_2 & \mathbf B^T \\ \mathbf B & \mathbf C_2 \end{bmatrix}}_{\mathbf M_2}$$

$$\mathbf M$$ and $$\mathbf M_1$$ are symmetric and positive definite. $$\mathbf M_2$$ are symmetric but not necessarily positive definite. However, $$\mathbf M_2$$ as well as $$\mathbf M$$ and $$\mathbf M_1$$ has all of its diagonal elements to be positive (while each off-diagonal element are free to be positive or negative).

My question is whether it is possible to prove that the diagonal elements in $$\mathbf M^{-1}$$ are smaller than their counterparts in $$\mathbf M_1^{-1}$$:

$$diag(\mathbf M^{-1})_{ii} \leq diag(\mathbf M_1^{-1})_{ii}$$

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More information: the size of $$\mathbf A$$, $$\mathbf A_1$$ and $$\mathbf A_2$$ is $$h \times h$$; the size of $$\mathbf C$$, $$\mathbf C_1$$ and $$\mathbf C_2$$ is $$k \times k$$; the size of $$\mathbf B$$ and $$\mathbf 0$$ is $$k \times h$$.

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I have tried several ways to this deal with question, such as Block inversion, Woodbury matrix identity, Schur complement..., but fail get a result, so I am here to seek your help. Thank you very much.

I come out with a limited solution by using Sherman–Morrison formula. The solution relies on an additional assumption that $$\mathbf M_2 = \mathbf m_2 \mathbf m_2^T$$, where $$\mathbf m_2$$ is a $$(h+k) \times 1$$ column vector. Given this assumption and by the definition of $$\mathbf M$$, we can express $$\mathbf M^{-1}$$ as follows:

$$\mathbf M^{-1} = (\mathbf M_1 + \mathbf M_2)^{-1} = (\mathbf M_1 + \mathbf m_2 \mathbf m_2^T)^{-1} = \mathbf M_1^{-1} - \frac{ \mathbf M_1^{-1} \mathbf m_2 \mathbf m_2^T \mathbf M_1^{-1}} {1 + \mathbf m_2^T \mathbf M_1^{-1} \mathbf m_2 }$$

where:

(1) $$\mathbf m_2^T \mathbf M_1^{-1} \mathbf m_2 >0$$, because, by definition, $$\mathbf M_1$$ is a positive definite.

(2) $$diag(\mathbf M_1^{-1} \mathbf m_2 \mathbf m_2^T \mathbf M_1^{-1})_{ii} >0$$, because $$\mathbf M_1^{-1} \mathbf m_2 \mathbf m_2^T \mathbf M_1^{-1} = (\mathbf M_1^{-1} \mathbf m_2) (\mathbf M_1^{-1} \mathbf m_2)^T$$.

(3) $$diag(\mathbf M^{-1})_{ii}>0$$ and $$diag(\mathbf M_1^{-1})_{ii}>0$$, because, by definition, both $$\mathbf M$$ and $$\mathbf M_1$$ are positive definite.

Taken together, given (1), (2) and (3), we have:

$$diag(\mathbf M^{-1})_{ii} < diag(\mathbf M_1^{-1})_{ii}$$

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However, I still wonder whether there is a more general proof for the above inequality. Is it possible to relax the assumption $$\mathbf M_2 = \mathbf m_2 \mathbf m_2^T$$ and prove the inequality?

• I'm joining this discussion a bit late, but I think you can extend that argument to any diagonizable matrix $M_2$ by writing $M_2 = P D P^T$ and then expand $D$ into a sum of outer products of scaled column vectors. You can then proceed iteratively to first show that diag(M)<diag(M_1^{-1} + all but one outer product) and keep on going. I'd like to add that there is a physics application for that - it shows that adding the results of measurements to previous measurements reduce the diagonal elements of the covariance matrix. Sep 11, 2021 at 16:41