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Let $R=\begin{bmatrix} R^{11} & R^{12} & R^{13} \\ R^{21} & R^{22} & R^{23}\\ R^{31} & R^{32} & R^{33} \end{bmatrix}$ be a symmetric positive definite matrix where $R^{ii}$, $i=1,2,3$ are square matrices. Since R is positive definite, matrices $R^{11}$, $R^{22}$,$R^{33}$ are also positive definite. My question is that: what can we say about matrix $A = R^{11} - R^{13}(R^{33})^{-1}R^{31}$? Positive definite, positive semi-definite or ...?

m0_as
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1 Answers1

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Short Answer: $R^{11}$, $R^{22}$, $R^{33}$ and $A$ are all positive definite, due to the relation between positive definiteness of a matrix and it's Schur-complement.

Longer Answer:Split $x$ into $\left[ x^T_1,x^T_2,x^T_3 \right]^T$. Now notice that condition $x^TRx > 0, \forall x \neq 0$ implies also $x^{T}_{j} R_{jj} x_j > 0, \forall x_j \neq 0$, simply by putting the corresponding subvectors of $x$ to zero. To reason about $A$, do the following: If $R$ is positive definite, then equivalently $\hat{R} = \Pi^\mathrm{T} R \Pi$ is positive definite for $\Pi$ being a permutation matrix such that

$\hat{R} = \begin{bmatrix} R_{11} & R_{13} & R_{12}\\ R_{31}& R_{33} & R_{23}\\ R_{21} & R_{32} & R_{22} \end{bmatrix}$

Now notice that with a similar argument as before the submatrix $\begin{bmatrix} R_{11} & R_{13} \\ R_{31}& R_{33} \end{bmatrix}$ is positive definite and therefore using the theorem about schur complements, it's schur complement with respect to $R_{33}$, (which we also showed is positive definite) which is $R_{11}-R_{13}R^{-1}_{33}R_{31}$ is also positive definite.

Dimitar Ho
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  • $A$ is not the Schur-Complement of any submatrix of R. If we had, $\tilde{R} = \begin{bmatrix} R^{11} & R^{13} \ R^{31} & R^{33} \end{bmatrix}$, then $A$ was the Schur-complement of $R^{33}$ of $\tilde{R}$. – m0_as Apr 28 '16 at 23:33
  • Oh, correct, first we can show that $\tilde{R}$ is positive definite, and then, the positive definiteness of $A$ is concluded :) – m0_as Apr 28 '16 at 23:36