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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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Eh, that is bad. Is there any straightforward assumption that could make the claim true? – Henrik Apr 10 '12 at 15:46
I am a bit confused. What does $[x \; \iota]$ mean? Since both $x$ and $\iota$ are $n \times 1$, standard matrix multiplication between them is not defined. – Calle Apr 10 '12 at 16:04
I've added how i thought the matrix should look. – Henrik Apr 10 '12 at 16:21
Thanks. So your result will be a new matrix. So you mean that $B$ will be positive definite if all the entries in this new matrix are positive, for all $x$? Or something else? – Calle Apr 10 '12 at 16:26
Well x is a given vector (which we can assume positive if needed). And I need to prove that B is positive definite (it's just a normal 2x2, right?). Does that answer your question? – Henrik Apr 10 '12 at 16:33
up vote 2 down vote accepted

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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You presumed right! I can show that linearly independency is a sufficient condition - but is it also a necessary condition? – Henrik Apr 10 '12 at 18:22
If they are linearly dependent then you can find a non zero $y$ such that $[x\;\iota] y = 0$. Then $y^T B y = 0$, so $B$ is only semi-definite. – copper.hat Apr 10 '12 at 19:34
Thank you very much! – Henrik Apr 10 '12 at 19:35
I know this is an old question, but the given answer is wrong. For any $x$ and $\iota$, $\det[x\iota]=0$, because for a vector $y$, $[x\iota]y = (\iota\cdot y)x$, and obviously there are vectors orthogonal to $\iota$ as long as $n>1$. So $B$ can be at best semi-definite. – Rhys Apr 25 '13 at 10:01
@Rhys: How is $\det \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = 0$??? Also, just from the dimensions, we must have $n=2$. – copper.hat Apr 25 '13 at 14:55

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