# Diagonal approximation of the inverse of a sparse matrix

Given a large sparse matrix $\Sigma$, I want to find a (block) diagonal matrix $\Omega$ which approximates $\Sigma^{-1}$.

In this specific problem, we know that $\Sigma$ is a block tridiagonal matrix. The blocks are of size $3 \times 3$. The blocks on the diagonal of $\Sigma$ stem from covariance matrices, i.e. they are symmetric positive semi-definite. The block $\Sigma_{i,i}$ represents the measurement uncertainty of the parameter block $i$ in my nonlinear least squares problem. The blocks $\Sigma_{i,i+1}$ on the upper diagonal represent the covariance matrix between parameter $i$ and $i+1$. (In general, they are smaller than the blocks on the diagonal, if that helps.) Therefore, $\Sigma$ is symmetric, positive semi-definite, and contains only real values.

As stated above, I am interested in finding the blockdiagonal matrix $\Omega$ which (best) approximates $\Sigma^{-1}$ in the sense that $\Omega \cdot \Sigma \approx I$. In the literature, I've often seen that the off-diagonal entries of $\Sigma$ are ignored which makes it easy to construct $\Omega$. I'd think there should be better approximations for finding $\Omega$.

I've also found this question, which is interested in the diagonal of the exact inverse. Again, I'd intuitively expect that this is not optimal as the off-diagonal entries of $\Omega$ are not small compared to its diagonal entries. This question aims to find any sparse matrix that approximates the inverse, but I am specifically interested in an approximation of a block diagonal form.

What is the best way to approach this problem?

• Are you aware that, most often, the inverse of a tridiagonal matrix can be far from sparse ? In particular far from a tridiagonal matrix (se the example of matrices $M$ and $D$ I give in one of my questions (math.stackexchange.com/q/1774670)). – Jean Marie Sep 3 '16 at 10:25
• Yes, I am aware of that. – stereo Sep 3 '16 at 10:34
• Why do you expect, that there exist a block diagonal $\Omega$, such that $\Omega\Sigma\approx I$? If this is the case, then $\Sigma$ is close to block diagonal and you can ignore off-diagonals. There are other methods, but they works only for some $\Sigma$ and in general can produce worse approximation. – Pawel Kowal Sep 3 '16 at 11:49
• Actually, I am not sure that such a matrix $\Omega$ exists. If you can make a strong argument that for my specific problem such a matrix a) does not exist, or b) can not be constructed efficiently, then I'd be satisfied and would drop this problem. – stereo Sep 3 '16 at 12:10

## 1 Answer

If $\Omega\Sigma = I + E$, with $\|E\|$ small, then $\Sigma^{-1}\Omega^{-1} \approx I - E$, and $\Sigma-\Omega^{-1} \approx \Sigma E$, thus $\|\Sigma-\Omega^{-1}\| \approx \|\Sigma\| \|E\|$.

Since $\Omega^{-1}$ is block diagonal, this implies, that norm of off-diagonal elements of $\Sigma$, $\Sigma_{off}$ is small: $\|\Sigma_{off}\|_1 \leq \|\Sigma\|_1 \|E\|_1$. This follows from: $\|X\|_1 \geq |X_{off}\|_1$ for any matrix $X$, where $X_{off}$ is a matrix constructed from $X$ by zeroing elements on block diagonals.

In your case off-diagonal entries of $\Sigma$ are not small compared to its diagonal elements, therefore you cannot find block diagonal $\Omega$, such that $\Omega\Sigma \approx I$.