1
$\begingroup$

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 $\mathbf M_1^{-1}$:

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

--

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 $k \times h$.

--

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.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

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:

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

--

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Sep 11, 2021 at 16:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .