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Suppose $\Sigma$ is a real-valued covariance matrix partitioned into equal-sized blocks as follows $$ \left(\begin{matrix} A&C\\ C^T&B \end{matrix}\right) = \Sigma $$

The following matrices are not symmetric yet seem to always have real eigenvalues, can anyone see a way to prove this?

$$D_1=C^{-1}B(C^{-1}B+(AC)^{-1})^{-1}$$ $$D_2=(BC^{-T}+CA^{-1})^{-1}CA^{-1}$$

Yaroslav Bulatov
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  • One things to check is whether these matrices happen to be similar to something positive definite. – Ben Grossmann Nov 13 '20 at 22:49
  • @BenGrossmann btw, recently saw that a product of $k$ random matrices will be not symmetric, but will have real eigenvalues for large $k$. Is there an intuitive understanding of signfiicance of non-symmetric matrix having real eigenvalues? https://arxiv.org/abs/1504.06256 – Yaroslav Bulatov Jul 18 '23 at 19:45
  • Interesting. My instinct is to think about this in terms of the characteristic polynomial: in principle, the characteristic polynomial of a non-symmetric matrix (with real entries) can be any polynomial (with real coefficients). So, the question becomes: what structure is being imposed on this polynomial that is causing it to have real roots? – Ben Grossmann Jul 18 '23 at 19:52

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Note that $D_1$ has real eigenvalues iff $D_1^{-1}$ has real eigenvalues. We find that $$ D_1^{-1} = (C^{-1}B + (AC)^{-1})B^{-1}C = C^{-1}BB^{-1}C + C^{-1}A^{-1}B^{-1}C\\ = I + C^{-1}A^{-1}B^{-1}C. $$ So, it suffices to show that $D_1^{-1} - I = C^{-1}A^{-1}B^{-1}C$ always has real eigenvalues. We note that this matrix is similar to $$ C(D_1^{-1} - I)C^{-1} = C[C^{-1}A^{-1}B^{-1}C]C^{-1} = A^{-1}B^{-1}, $$ which is in turn similar to $$ A^{1/2}A^{-1}B^{-1}A^{-1/2} = A^{-1/2}B^{-1}A^{-1/2}. $$ So, we see that $D_1^{-1} - I$ is similar to $A^{-1/2}B^{-1}A^{-1/2}$. $B$ is a principal submatrix of a positive definite matrix and is therefore positive definite, which means that $B^{-1}$ is positive definite, which means that $A^{-1/2}B^{-1}A^{-1/2} = [A^{-1/2}]^TB^{-1}A^{-1/2}$ is positive definite, which means that $A^{-1/2}B^{-1}A^{-1/2}$ has real eigenvalues, which means that $D_1^{-1} - I$ has real eigenvalues.

A similar line of reasoning can be applied to $D_2$.

Ben Grossmann
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  • You seem to imply that $A^{-1/2}\left(AC^{-1}B^{-1}C\right)A^{-1/2}$ is equal or similar to $\left(AC^{-1}B^{-1}C\right)$. Is that so? I would have thought it ought to be $A^{-1/2}\left(AC^{-1}B^{-1}C\right)A^{+1/2}$, or something like it – Cryo Nov 13 '20 at 23:18
  • @Cryo you’re right, that was a careless error. Not sure if I can fix this answer – Ben Grossmann Nov 14 '20 at 00:24
  • There's a typo in first line -- $(AC)^{-1}$ became $AC^{-1}$ – Yaroslav Bulatov Nov 14 '20 at 00:51
  • @YaroslavBulatov That was the big mistake! Good catch. See my latest edit; I think the same argument works now. – Ben Grossmann Nov 14 '20 at 01:09
  • Assuming I didn't make any mistakes this time, this all means that $[A^{1/2}C]D_1[A^{1/2}C]^{-1}$ is positive definite, which gives you an alternate (perhaps quicker) proof. – Ben Grossmann Nov 14 '20 at 01:12