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Given the positive vector ${\bf a} \in {\Bbb R}_{>0}^n$ and symmetric positive definite $n \times n$ matrices ${\bf A}_1, {\bf A}_2, \dots, {\bf A}_n$, define the scalar field $f : {\Bbb R}^n \to {\Bbb R}$ by

$$ f({\bf x}) := ({\bf x} - {\bf a})^\top \left( \sum_{i=1}^n x_i {\bf A}_i \right)^{-1}({\bf x} - {\bf a}) $$

$x_i \geq a_i$ for all $i$. Is $f$ convex?

Without the $x_i$ in the inner parenthesis the answer is yes. I have tried to look at the shape of $f$ numerically and it does look convex for the parameters I have tried.

Rodrigo de Azevedo
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user_lambda
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  • I suspect you do not know what a quadratic form is. A quadratic form is merely a quadratic polynomial in which every monomial is degree-2. Something like vector-transpose-times-matrix-times-vector is not necessarily a quadratic form – Rodrigo de Azevedo Jul 21 '23 at 05:12
  • Yes thanks for clarifying the text. – user_lambda Jul 21 '23 at 12:50

2 Answers2

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Introducing a new variable, $y \in {\Bbb R}$, the epigraph of $f$ is defined by

$$ ({\bf x} - {\bf a})^\top \left( \sum_{i=1}^n x_i {\bf A}_i \right)^{-1}({\bf x} - {\bf a}) \leq y $$

which, via the Schur complement, can be rewritten as the following linear matrix inequality (LMI).

$$ \begin{bmatrix} \displaystyle\sum_{i=1}^n x_i {\bf A}_i & ({\bf x} - {\bf a}) \\ ({\bf x} - {\bf a})^\top & y\end{bmatrix} \succeq {\bf O}_{n+1} $$

Hence, the epigraph of $f$ is a spectrahedron and, thus, convex. Therefore, $f$ is also convex.

Related

Rodrigo de Azevedo
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  • Thanks for your response. Could you explain how we can rewrite the epigraph inequality into the linear matrix inequality? I've looked at the wikipedia article for the Schur complement but I still don't see how that step works. – user_lambda Jul 21 '23 at 13:59
  • @user_lambda Try to understand this first. Do you understand every step? This is basically Gaussian elimination for block matrices. – Rodrigo de Azevedo Jul 21 '23 at 14:01
  • Yes I understand every step in that response. – user_lambda Jul 21 '23 at 14:46
  • @user_lambda Then you can apply it to this answer. Start from the bottom and try to compute the Schur complement of the northwestern block – Rodrigo de Azevedo Jul 21 '23 at 14:48
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    I see. So the Schur complement of the northwestern block is $$ y-({\bf x} - {\bf a})^\top \left( \sum_{i=1}^n x_i {\bf A}_i \right)^{-1}({\bf x} - {\bf a}).$$ Then we know that the bottom matrix is PSD iif the Schur complement is PSD. But since the Schur complement is a scalar we have a normal inequality. Very clever! Thanks a lot for your help. – user_lambda Jul 21 '23 at 15:11
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    @user_lambda You are welcome. That is correct. You may want to take a look at Jean Gallier's notes. The relevant result is this one – Rodrigo de Azevedo Jul 21 '23 at 15:32
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Let $B \equiv \sum_i x_i A_i$. Since $A_i$ are each symmetric and positive definite and $x_i > 0$, then $B$ is also symmetric and positive definite. $$ B[j,k] = \sum_i x_i A_i[j,k] = \sum_i x_i A_i[k,j] = B[k,j] \\ y^T B y = \sum_i x_i y^T A_i y > 0 $$

Consider some $y,z$ such that $By=z$. Then we can show that $B^{-1}$ is also symmetric and positive definite. $$ B^{-1} = B^{-1} I = B^{-1} I^T = B^{-1} (B^{-1} B)^T = B^{-1} B^T (B^{-1})^T = B^{-1} B (B^{-1})^T = (B^{-1})^T\\ z^T B^{-1}z = z^Ty = (By)^T y = y^T B y > 0 $$

Now define $z \equiv x - a$. Then, $$f(x) = g(z) = z^T B^{-1} z$$

So $g$ is convex in $z$. Since $f$ has the same value, but whose input is shifted by a constant $a$, $f$ is also convex in $x$.

Tony Mathew
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    This is not correct. You are suppressing that $A$ (or, equivalently $B$) also depends on $x$ (or, equivalently, $z$). – gerw Jul 21 '23 at 12:41