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Okay, so this hopefully is an easy question, but I'm not that much into linear algebra. Could someone help me realize the following:

A is symmetric, positive definite $n\times n$, x is $n\times 1$ and non-negative (if we can relax that assumption it would be great), $\iota\text{ is }n\times 1$ and only consists of 1's.

I wan't $B=[x\;\iota]^{T}A^{-1}[x\;\iota]=\left( \begin{array}{cc} x^TA^{-1}x & x^TA^{-1}\iota\\ \iota^T A^{-1}x & \iota^T A^{-1}\iota \\ \end{array} \right)$ to be positive definite. How can I show that?

Best regards, Henrik

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    Oh right, now I get it.2012-04-10

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I presume $B \in \mathbb{R}^{2\times 2}$, otherwise I have misunderstood.

If $A$ is symmetric and positive definite, then so is $A^{-1}$.

In order that $B$ be positive definite, you need $[x\;\iota]$ to be to have a trivial null space, so you just need $x, \iota$ to be linearly independent.

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    Sorry, I just looked at the written-out matrix. I assumed that $[x\iota]$ was supposed to represent the outer product, written as a matrix (so that $B$ is also $n\times n$).2013-04-25