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After writing up some math, I ended up with a term like so:

$\left(A^{-1} + B^{-1}\right)^{-1}$

where $A$ and $B$ are 2 covariance matrices.

1) Can I be sure that this expression is meaningful? (i.e. $\left(A^{-1} + B^{-1}\right)$ is not singular)

2) If $A$ and $B$ were numbers, I could rewrite this as $\frac{AB}{A+B}$. Is there a similar formula valid for matrices? Ultimately I wonder if this expression can be simplified: $z^T\left(A^{-1} + B^{-1}\right)^{-1}z$
where $z=A^{-1}x + B^{-1}y$ , for some vectors $x$ and $y$.

Julien
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1 Answers1

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  1. If $A$ and $B$ are invertible covariance matrices, they must be positive definite; hence their inverses must be positive definite. But the sum of two positive definite matrices is again positive definite. (Proof: when $u\neq0$, $u^T(A+B)u=u^tAu+u^tBu>0+0=0$). Therefore $A^{-1}+B^{-1}$ is positive definite, hence invertible.

  2. See Inverse of the sum of matrices. Note your formula has at least four different interpretations for matrices: $AB(A+B)^{-1}$, $(A+B)^{-1}AB$, etc.

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symplectomorphic
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  • Of course! I didn't attack this the right way. For 2): Yes I'm aware that this notation is ambiguous for matrices and also likely wrong, that's why I ask for an analog formula that would be valid ;) – Julien Aug 10 '16 at 05:48
  • @JulienBernu: see my update. Note, by the way, that $ab/(a+b)$ is not quite the harmonic mean. You're missing a factor of two. – symplectomorphic Aug 10 '16 at 05:52
  • Had a look at the link but did not find much useful. See my edit for 2) though, maybe you'll see something. Or maybe there's just nothing to simplify... Thanks! – Julien Aug 10 '16 at 06:17